Stalk of a pushforward sheaf in algebraic geometry

No, I'm afraid what you say does not work, even in the simplest situation:

a) Let $k$ be a field , $X=\mathbb A^1_k=Spec (k[t])$ , $Y=Spec (k)=\lbrace y\rbrace$ and let $f:X\to Y$ be the obvious morphism.
Then for $\mathcal F=\mathcal O_X$ we have $(f_*\mathcal F)_y=k[t]$.

b) On the other hand, still in your notation, $X=Z_1$, $\mathfrak p_1=(0)$, $M=k[t]$, $S=k[t]\setminus \lbrace 0\rbrace $ and thus $S^{-1}M=k(t)$

c) Conclusion: $$(f_*\mathcal F)_y=k[t]\neq S^{-1}M=k(t)$$

I think the best we can do is the following. The question is local on the base, so we may as well assume $Y = \text{Spec } A$ and $X = \text{Spec } B$ are affine and $\mathcal{F} = \widetilde{M}$ where $M$ is a $B$-module. Then if $y = \mathfrak{p} \in \text{Spec } A$ then $(f_*\mathcal{F})_y$ is the $A_{\mathfrak{p}}$-module $M_{\mathfrak{p}}$, where we think of $M$ now as an $A$-module.