The assumptions on $X$ and $Y$ mean that $f$ is the base change of $f'$ by $X \to X'$. The properties $\mathcal{P}$ listed in (1) -- (23) above are all stable under base change, hence if $f'$ has property $\mathcal{P}$, then so does $f$. See Schemes, Lemmas \ref{schemes-lemma-base-change-immersion}, \ref{schemes-lemma-quasi-compact-preserved-base-change}, \ref{schemes-lemma-separated-permanence}, and \ref{schemes-lemma-base-change-monomorphism} and Morphisms, Lemmas \ref{morphisms-lemma-base-change-surjective}, \ref{morphisms-lemma-base-change-universally-injective}, \ref{morphisms-lemma-base-change-affine}, \ref{morphisms-lemma-base-change-finite-type}, \ref{morphisms-lemma-base-change-quasi-finite}, \ref{morphisms-lemma-base-change-finite-presentation}, \ref{morphisms-lemma-base-change-relative-dimension-d}, \ref{morphisms-lemma-base-change-syntomic}, \ref{morphisms-lemma-base-change-smooth}, \ref{morphisms-lemma-base-change-unramified}, \ref{morphisms-lemma-base-change-etale}, \ref{morphisms-lemma-base-change-proper}, \ref{morphisms-lemma-base-change-finite}, and \ref{morphisms-lemma-base-change-finite-locally-free}. \medskip\noindent The interesting direction in each case is therefore to assume that $f$ has the property and deduce that $f'$ has it too. By induction on the order of the thickening we may assume that $S \subset S'$ is a first order thickening, see discussion immediately following Definition \ref{definition-thickening}. We make a couple of general remarks which we will use without further mention in the arguments below. (I) Let $W' \subset S'$ be an affine open and let $U' \subset X'$ and $V' \subset Y'$ be affine opens lying over $W'$ with $f'(U') \subset V'$. Let $W' = \Spec(R')$ and denote $I \subset R'$ be the ideal defining the closed subscheme $W' \cap S$. Say $U' = \Spec(B')$ and $V' = \Spec(A')$. Then we get a commutative diagram $$ \xymatrix{ 0 \ar[r] & IB' \ar[r] & B' \ar[r] & B \ar[r] & 0 \\ 0 \ar[r] & IA' \ar[r] \ar[u] & A' \ar[r] \ar[u] & A \ar[r] \ar[u] & 0 } $$ with exact rows. Moreover $IB' \cong I \otimes_R B$, see proof of Lemma \ref{lemma-deform}. (II) The morphisms $X \to X'$ and $Y \to Y'$ are universal homeomorphisms. Hence the topology of the maps $f$ and $f'$ (after any base change) is identical. (III) If $f$ is flat, then $f'$ is flat and $Y' \to S'$ is flat at every point in the image of $f'$, see Lemma \ref{lemma-flatness-morphism-thickenings}. \medskip\noindent Ad (\ref{item-flat}). This is general remark (III). \medskip\noindent Ad (\ref{item-isomorphism}). Assume $f$ is an isomorphism. By (III) we see that $Y' \to S'$ is flat. Choose an affine open $V' \subset Y'$ and set $U' = (f')^{-1}(V')$. Then $V = Y \cap V'$ is affine which implies that $V \cong f^{-1}(V) = U = Y \times_{Y'} U'$ is affine. By Lemma \ref{lemma-thickening-affine-scheme} we see that $U'$ is affine. Thus we have a diagram as in the general remark (I) and moreover $IA \cong I \otimes_R A$ because $R' \to A'$ is flat. Then $IB' \cong I \otimes_R B \cong I \otimes_R A \cong IA'$ and $A \cong B$. By the exactness of the rows in the diagram above we see that $A' \cong B'$, i.e., $U' \cong V'$. Thus $f'$ is an isomorphism. \medskip\noindent Ad (\ref{item-open-immersion}). Assume $f$ is an open immersion. Then $f$ is an isomorphism of $X$ with an open subscheme $V \subset Y$. Let $V' \subset Y'$ be the open subscheme whose underlying topological space is $V$. Then $f'$ is a map from $X'$ to $V'$ which is an isomorphism by (\ref{item-isomorphism}). Hence $f'$ is an open immersion. \medskip\noindent Ad (\ref{item-quasi-compact}). Immediate from remark (II). See also Lemma \ref{lemma-thicken-property-morphisms} for a more general statement. \medskip\noindent Ad (\ref{item-universally-closed}). Immediate from remark (II). See also Lemma \ref{lemma-thicken-property-morphisms} for a more general statement. \medskip\noindent Ad (\ref{item-separated}). Note that $X \times_Y X = Y \times_{Y'} (X' \times_{Y'} X')$ so that $X' \times_{Y'} X'$ is a thickening of $X \times_Y X$. Hence the topology of the maps $\Delta_{X/Y}$ and $\Delta_{X'/Y'}$ matches and we win. See also Lemma \ref{lemma-thicken-property-morphisms} for a more general statement. \medskip\noindent Ad (\ref{item-monomorphism}). Assume $f$ is a monomorphism. Consider the diagonal morphism $\Delta_{X'/Y'} : X' \to X' \times_{Y'} X'$. The base change of $\Delta_{X'/Y'}$ by $S \to S'$ is $\Delta_{X/Y}$ which is an isomorphism by assumption. By (\ref{item-isomorphism}) we conclude that $\Delta_{X'/Y'}$ is an isomorphism. \medskip\noindent Ad (\ref{item-surjective}). This is clear. See also Lemma \ref{lemma-thicken-property-morphisms} for a more general statement. \medskip\noindent Ad (\ref{item-universally-injective}). Immediate from remark (II). See also Lemma \ref{lemma-thicken-property-morphisms} for a more general statement. \medskip\noindent Ad (\ref{item-affine}). Assume $f$ is affine.  Choose an affine open $V' \subset Y'$ and set $U' = (f')^{-1}(V')$. Then $V = Y \cap V'$ is affine which implies that $U = Y \times_{Y'} U'$ is affine. By Lemma \ref{lemma-thickening-affine-scheme} we see that $U'$ is affine. Hence $f'$ is affine. See also Lemma \ref{lemma-thicken-property-morphisms} for a more general statement. \medskip\noindent Ad (\ref{item-finite-type}). Via remark (I) comes down to proving $A' \to B'$ is of finite type if $A \to B$ is of finite type. Suppose that $x_1, \ldots, x_n \in B'$ are elements whose images in $B$ generate $B$ as an $A$-algebra. Then $A'[x_1, \ldots, x_n] \to B$ is surjective as both $A'[x_1, \ldots, x_n] \to B$ is surjective and $I \otimes_R A[x_1, \ldots, x_n] \to I \otimes_R B$ is surjective. See also Lemma \ref{lemma-thicken-property-morphisms-cartesian} for a more general statement. \medskip\noindent Ad (\ref{item-quasi-finite}). Follows from (\ref{item-finite-type}) and that quasi-finiteness of a morphism of finite type can be checked on fibres, see Morphisms, Lemma \ref{morphisms-lemma-quasi-finite-at-point-characterize}. See also Lemma \ref{lemma-thicken-property-morphisms-cartesian} for a more general statement. \medskip\noindent Ad (\ref{item-finite-presentation}). Via remark (I) comes down to proving $A' \to B'$ is of finite presentation if $A \to B$ is of finite presentation. We may assume that $B' = A'[x_1, \ldots, x_n]/K'$ for some ideal $K'$ by (\ref{item-finite-type}). We get a short exact sequence $$ 0 \to K' \to A'[x_1, \ldots, x_n] \to B' \to 0 $$ As $B'$ is flat over $R'$ we see that $K' \otimes_{R'} R$ is the kernel of the surjection $A[x_1, \ldots, x_n] \to B$. By assumption on $A \to B$ there exist finitely many $f'_1, \ldots, f'_m \in K'$ whose images in $A[x_1, \ldots, x_n]$ generate this kernel. Since $I$ is nilpotent we see that $f'_1, \ldots, f'_m$ generate $K'$ by Nakayama's lemma, see Algebra, Lemma \ref{algebra-lemma-NAK}. \medskip\noindent Ad (\ref{item-relative-dimension-d}). Follows from (\ref{item-finite-type}) and general remark (II). See also Lemma \ref{lemma-thicken-property-morphisms-cartesian} for a more general statement. \medskip\noindent Ad (\ref{item-universally-open}). Immediate from general remark (II). See also Lemma \ref{lemma-thicken-property-morphisms} for a more general statement. \medskip\noindent Ad (\ref{item-syntomic}). Assume $f$ is syntomic. By (\ref{item-finite-presentation}) $f'$ is locally of finite presentation, by general remark (III) $f'$ is flat and the fibres of $f'$ are the fibres of $f$. Hence $f'$ is syntomic by Morphisms, Lemma \ref{morphisms-lemma-syntomic-flat-fibres}. \medskip\noindent Ad (\ref{item-smooth}). Assume $f$ is smooth. By (\ref{item-finite-presentation}) $f'$ is locally of finite presentation, by general remark (III) $f'$ is flat, and the fibres of $f'$ are the fibres of $f$. Hence $f'$ is smooth by Morphisms, Lemma \ref{morphisms-lemma-smooth-flat-smooth-fibres}. \medskip\noindent Ad (\ref{item-unramified}). Assume $f$ unramified. By (\ref{item-finite-type}) $f'$ is locally of finite type and the fibres of $f'$ are the fibres of $f$. Hence $f'$ is unramified by Morphisms, Lemma \ref{morphisms-lemma-unramified-etale-fibres}. See also Lemma \ref{lemma-thicken-property-morphisms-cartesian} for a more general statement. \medskip\noindent Ad (\ref{item-etale}). Assume $f$ \'etale. By (\ref{item-finite-presentation}) $f'$ is locally of finite presentation, by general remark (III) $f'$ is flat, and the fibres of $f'$ are the fibres of $f$. Hence $f'$ is \'etale by Morphisms, Lemma \ref{morphisms-lemma-etale-flat-etale-fibres}. \medskip\noindent Ad (\ref{item-proper}). This follows from a combination of (\ref{item-separated}), (\ref{item-finite-type}), (\ref{item-quasi-compact}), and (\ref{item-universally-closed}).  See also Lemma \ref{lemma-thicken-property-morphisms-cartesian} for a more general statement. \medskip\noindent Ad (\ref{item-integral}). Combine (\ref{item-universally-closed}) and (\ref{item-affine}) with Morphisms, Lemma \ref{morphisms-lemma-integral-universally-closed}. See also Lemma \ref{lemma-thicken-property-morphisms} for a more general statement. \medskip\noindent Ad (\ref{item-finite}). Combine (\ref{item-integral}), and (\ref{item-finite-type}) with Morphisms, Lemma \ref{morphisms-lemma-finite-integral}. See also Lemma \ref{lemma-thicken-property-morphisms-cartesian} for a more general statement. \medskip\noindent Ad (\ref{item-finite-locally-free}). Assume $f$ finite locally free. By (\ref{item-finite}) we see that $f'$ is finite, by general remark (III) $f'$ is flat, and by (\ref{item-finite-presentation}) $f'$ is locally of finite presentation. Hence $f'$ is finite locally free by Morphisms, Lemma \ref{morphisms-lemma-finite-flat}.