The properties $\mathcal{P}$ listed in the lemma 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} and \ref{schemes-lemma-base-change-monomorphism} and Morphisms, Lemmas \ref{morphisms-lemma-base-change-finite-type}, \ref{morphisms-lemma-base-change-quasi-finite}, \ref{morphisms-lemma-base-change-relative-dimension-d}, \ref{morphisms-lemma-base-change-differentials}, \ref{morphisms-lemma-base-change-unramified}, \ref{morphisms-lemma-base-change-proper}, and \ref{morphisms-lemma-base-change-finite}. \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}. \medskip\noindent Most of the proofs will use a reduction to the affine case. Let $U' \subset S'$ be an affine open and let $V' \subset X'$ be an affine open lying over $U'$. Let $U' = \Spec(A')$ and denote $I \subset A'$ be the ideal defining the closed subscheme $U' \cap S$. Say $V' = \Spec(B')$. Then $V' \cap X = \Spec(B'/IB')$. Setting $A = A'/I$ and $B = B'/IB'$ 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 and $I^2 = 0$. \medskip\noindent The translation of (1) into algebra: If $A \to B$ is surjective, then $A' \to B'$ is surjective. This follows from Nakayama's lemma (Algebra, Lemma \ref{algebra-lemma-NAK}). \medskip\noindent The translation of (2) into algebra: If $A \to B$ is a finite type ring map, then $A' \to B'$ is a finite type ring map. This follows from Nakayama's lemma (Algebra, Lemma \ref{algebra-lemma-NAK}) applied to a map $A'[x_1, \ldots, x_n] \to B'$ such that $A[x_1, \ldots, x_n] \to B$ is surjective. \medskip\noindent Proof of (3).  Follows from (2) and that quasi-finiteness of a morphism which is locally of finite type can be checked on fibres, see Morphisms, Lemma \ref{morphisms-lemma-quasi-finite-at-point-characterize}. \medskip\noindent Proof of (4). Follows from (2) and that the additional property of ``being of relative dimension $d$'' can be checked on fibres (by definition, see Morphisms, Definition \ref{morphisms-definition-relative-dimension-d}. \medskip\noindent The translation of (5) into algebra: If $\Omega_{B/A} = 0$, then $\Omega_{B'/A'} = 0$. By Algebra, Lemma \ref{algebra-lemma-differentials-base-change} we have $0 = \Omega_{B/A} = \Omega_{B'/A'}/I\Omega_{B'/A'}$. Hence $\Omega_{B'/A'} = 0$ by Nakayama's lemma (Algebra, Lemma \ref{algebra-lemma-NAK}). \medskip\noindent The translation of (6) into algebra: If $A \to B$ is unramified map, then $A' \to B'$ is unramified. Since $A \to B$ is of finite type we see that $A' \to B'$ is of finite type by (2) above. Since $A \to B$ is unramified we have $\Omega_{B/A} = 0$. By part (5) we have $\Omega_{B'/A'} = 0$. Thus $A' \to B'$ is unramified. \medskip\noindent Proof of (7). Follows by combining (2) with results of Lemma \ref{lemma-thicken-property-morphisms} and the fact that proper equals quasi-compact $+$ separated $+$ locally of finite type $+$ universally closed. \medskip\noindent Proof of (8). Follows by combining (2) with results of Lemma \ref{lemma-thicken-property-morphisms} and using the fact that finite equals integral $+$ locally of finite type (Morphisms, Lemma \ref{morphisms-lemma-finite-integral}). \medskip\noindent Proof of (9). As $f$ is a monomorphism we have $X = X \times_S X$. We may apply the results proved so far to the morphism of thickenings $(X \subset X') \to (X \times_S X \subset X' \times_{S'} X')$. We conclude $X' \to X' \times_{S'} X'$ is a closed immersion by (1). In fact, it is a first order thickening as the ideal defining the closed immersion $X' \to X' \times_{S'} X'$ is contained in the pullback of the ideal $\mathcal{I} \subset \mathcal{O}_{S'}$ cutting out $S$ in $S'$. Indeed, $X = X \times_S X = (X' \times_{S'} X') \times_{S'} S$ is contained in $X'$. Hence by Morphisms, Lemma \ref{morphisms-lemma-differentials-diagonal} it suffices to show that $\Omega_{X'/S'} = 0$ which follows from (5) and the corresponding statement for $X/S$. \medskip\noindent Proof of (10). If $f : X \to S$ is an immersion, then it factors as $X \to U \to S$ where $U \to S$ is an open immersion and $X \to U$ is a closed immersion. Let $U' \subset S'$ be the open subscheme whose underlying topological space is the same as $U$. Then $X' \to S'$ factors through $U'$ and we conclude that $X' \to U'$ is a closed immersion by part (1). This finishes the proof.