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-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}. Hence in each case we need only to prove that if $f$ has the desired property, so does $f'$. \medskip\noindent A morphism is locally quasi-finite if and only if it is locally of finite type and the scheme theoretic fibres are discrete spaces, see Morphisms, Lemma \ref{morphisms-lemma-locally-quasi-finite-fibres}. Since the underlying topological space is unchanged by  passing to a thickening, we see that $f'$ is locally quasi-finite if (and only if) $f$ is. This proves (1). \medskip\noindent Case (2) follows from case (5) of Lemma \ref{lemma-thicken-property-morphisms} and the fact that the finite morphisms are precisely the integral morphisms that are locally of finite type (Morphisms, Lemma \ref{morphisms-lemma-finite-integral}). \medskip\noindent Case (3). This follows immediately from Morphisms, Lemma \ref{morphisms-lemma-check-closed-infinitesimally}. \medskip\noindent Case (4) follows from the following algebra statement: Let $A$ be a ring and let $I \subset A$ be a locally nilpotent ideal. Let $B$ be a finite type $A$-algebra. If $\Omega_{(B/IB)/(A/I)} = 0$, then $\Omega_{B/A} = 0$. Namely, the assumption means that $I\Omega_{B/A} = 0$, see Algebra, Lemma \ref{algebra-lemma-differentials-base-change}. On the other hand $\Omega_{B/A}$ is a finite $B$-module, see Algebra, Lemma \ref{algebra-lemma-differentials-finitely-generated}. Hence the vanishing of $\Omega_{B/A}$ follows from Nakayama's lemma (Algebra, Lemma \ref{algebra-lemma-NAK}) and the fact that $IB$ is contained in the Jacobson radical of $B$. \medskip\noindent Case (5) follows immediately from (4) and Morphisms, Lemma \ref{morphisms-lemma-unramified-omega-zero}. \medskip\noindent Proof of (6). As $f$ is a monomorphism we have $X = X \times_Y X$. We may apply the results proved so far to the morphism of thickenings $(X \subset X') \to (X \times_Y X \subset X' \times_{Y'} X')$. We conclude $\Delta_{X'/Y'} : X' \to X' \times_{Y'} X'$ is a closed immersion by (3). In fact $\Delta_{X'/Y'}$ is a bijection on underlying sets, hence $\Delta_{X'/Y'}$ is a thickening. On the other hand $\Delta_{X'/Y'}$ is locally of finite presentation by Morphisms, Lemma \ref{morphisms-lemma-diagonal-morphism-finite-type}. In other words, $\Delta_{X'/Y'}(X')$ is cut out by a quasi-coherent sheaf of ideals $\mathcal{J} \subset \mathcal{O}_{X' \times_{Y'} X'}$ of finite type. Since $\Omega_{X'/Y'} = 0$ by (5) we see that the conormal sheaf of $X' \to X' \times_{Y'} X'$ is zero by Morphisms, Lemma \ref{morphisms-lemma-differentials-diagonal}. In other words, $\mathcal{J}/\mathcal{J}^2 = 0$. This implies $\Delta_{X'/Y'}$ is an isomorphism, for example by Algebra, Lemma \ref{algebra-lemma-ideal-is-squared-union-connected}. \medskip\noindent Proof of (7). If $f : X \to Y$ is an immersion, then it factors as $X \to V \to Y$ where $V \to Y$ is an open immersion and $X \to V$ is a closed immersion. Let $V' \subset Y'$ be the open subscheme whose underlying topological space is the same as $V$. Then $X' \to V'$ factors through $V'$ and we conclude that $X' \to V'$ is a closed immersion by part (3). \medskip\noindent Case (8) follows from Lemma \ref{lemma-thicken-property-morphisms} and the definition of proper morphisms as being the quasi-compact, universally closed, and separated morphisms that are locally of finite type.