**Proof.**
Proof of (1). Choose a short exact sequence $0 \to K' \to (R')^{\oplus n} \to M' \to 0$. Let $K \subset R^{\oplus n}$ be the inverse image of $K'$ under the map $R^{\oplus n} \to (R')^{\oplus n}$. Then $M = R^{\oplus n}/K$ works.

Proof of (2). Choose a presentation $(R')^{\oplus m} \to (R')^{\oplus n} \to M' \to 0$. Suppose that the first map is given by the matrix $A' = (a'_{ij})$ and the second map is determined by generators $x'_ i \in M'$, $i = 1, \ldots , n$. As $R' = S^{-1}(R/I)$ we can choose $s \in S$ and a matrix $A = (a_{ij})$ with coefficients in $R$ such that $a'_{ij} = a_{ij} / s \bmod S^{-1}I$. Let $M$ be the finitely presented $R$-module with presentation $R^{\oplus m} \to R^{\oplus n} \to M \to 0$ where the first map is given by the matrix $A$ and the second map is determined by generators $x_ i \in M$, $i = 1, \ldots , n$. Then the map $M \to M'$, $x_ i \mapsto x'_ i$ induces an isomorphism $S^{-1}(M/IM) \cong M'$.
$\square$

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