First, if $h:A\rightarrow B$ is a morphism of rings, $M$ is a $B$-module, and $N\subseteq M$ is a sub-$B$-module, then it holds $M=N$ if and only if the underlying sets of $M$ and $N$ are equal, hence if and only if the $B$-modules obtained from $M$ and $N$ by scalar restriction along of $h$ are equal.

Second, if $A$ is a ring, $M$ is an $A$-module, and $N\subseteq M$ is a sub-$A$-module, then it holds $N=M$ if and only if $N_{\mathfrak{p}}=M_{\mathfrak{p}}$ for every prime ideal $\mathfrak{p}$ of $A$ (cf. [Bourbaki, Algèbre commutative, II.3.3 Théorème 1]).

Third, putting the above together answers what seems to be your first question.

Finally, it is not completely clear to me, what you want as an answer to your second question. Let me just point out that in the last paragraph of the proof one applies Lemma 15.1.4 (ii) and thus needs the base ring to be local with infinite residue field, and one moreover uses the facts that $R$ is *local and thus needs the base ring to be local.

First, if $h:A\rightarrow B$ is a morphism of rings, $M$ is a $B$-module, and $N\subseteq M$ is a sub-$B$-module, then it holds $M=N$ if and only if the underlying sets of $M$ and $N$ are equal, hence if and only if the $A$-modules obtained from $M$ and $N$ by scalar restriction along of $h$ are equal.

Second, if $A$ is a ring, $M$ is an $A$-module, and $N\subseteq M$ is a sub-$A$-module, then it holds $N=M$ if and only if $N_{\mathfrak{p}}=M_{\mathfrak{p}}$ for every prime ideal $\mathfrak{p}$ of $A$ (cf. [Bourbaki, Algèbre commutative, II.3.3 Théorème 1]).

Third, putting the above together answers what seems to be your first question.

Finally, it is not completely clear to me, what you want as an answer to your second question. Let me just point out that in the last paragraph of the proof one applies Lemma 15.1.4 (ii) and thus needs the base ring to be local with infinite residue field, and one moreover uses the facts that $R$ is *local and thus needs the base ring to be local.