You are starting with a sequence of $A$-modules
$$
0 \to M' \stackrel f\to M \stackrel g\to M'' \to 0.
$$
You want to show that if for each prime $\mathfrak{p}$ of $A$ the induced sequence of $A_\mathfrak{p}$-modules
$$
0 \to M'_\mathfrak{p} \stackrel{f_\mathfrak{p}}\longrightarrow M_\mathfrak{p} \stackrel{g_\mathfrak{p}}\longrightarrow M''_\mathfrak{p} \to 0
$$
is exact, then the original sequence is exact. Here the map $f_\mathfrak{p}$, for example, just sends $x/s$ to $f(x)/s$. You don't need to perform any sort of "lifting" on $f_\mathfrak{p}$, because you already have $f$. One way of solving the problem is to find ways to apply the fact that if $N_\mathfrak{p} = 0$ for all primes $\mathfrak{p}$, then $N = 0$. [You could try this with $N = \operatorname{Ker} f$. What is the relationship between $\operatorname{Ker} f$ and $\operatorname{Ker} f_\mathfrak{p}$?]
I should mention that if $M'$ is finitely presented then there is a relationship between $A$-linear $M' \to M$ and $S^{-1}A$-linear $S^{-1}M' \to S^{-1}M$. Proposition 2.10 of Eisenbud's book shows that in this situation we have a natural isomorphism
$$
\operatorname{Hom}_{S^{-1}A}(S^{-1}M', S^{-1}M) \approx S^{-1}\operatorname{Hom}_A(M', M).
$$
So you might suggestively call Georges' example "the identity map divided by $2$".