Let $\pi: X \to Y$ be a morphism of schemes mapping $x$ to $y$. Let $\mathscr{F}$ be a qcoh sheaf on $X$. Consider the fibre $X_y = \pi^{-1}(y)$, equipped with the natural morphism $i : X_y \to X$. The point $x$ is in this fibre. We can consider the pullback $i^* \mathscr{F}$ as a sheaf on $X_y$. I want to show $$ (i^*\mathscr{F})_x = \mathscr{F}_x/\mathfrak{m}_y \mathscr{F}_x. $$ In particular, when $\mathscr{F} = \mathcal{O}_X$ we get $\mathcal{O}_{X_y, x} = \mathcal{O}_{X,x}/\mathfrak{m}_y \mathcal{O}_{X,x}$. Here $\mathfrak{m}_y$ is the maximal ideal of the local ring $\mathcal{O}_{Y,y}$.
This result is used everywhere but I can't find a reference. I always get my algebra confused when I try to prove it. Here's what I have so far.
As we're looking at stalks we can work locally and assume our schemes are affine. This allows us to translate the above into the following algebraic setup. The map $\pi$ corresponds to a ring homomorphism $\varphi : B \to A$ with $\varphi^{-1}(\mathfrak{p}) = \mathfrak{q}$. The sheaf is an $A$-module $M$. The fibre is the (spectrum of) the ring $A \otimes_B k(y) = A_\mathfrak{q}/\mathfrak{q} A_\mathfrak{q}$. The inclusion of the point $x$ in the fibre $X_y$ corresponds to a homomorphism $\alpha : A \otimes_B k(y) \to k(x)$, and $x$ is the prime ideal $\mathfrak{p}' = \alpha^{-1}(0) \subset A \otimes_B k(y)$. Then what I want to show is $$ (A \otimes_B k(y))_{\mathfrak{p}'} = A_\mathfrak{p}/\mathfrak{q} A_\mathfrak{p}. $$ The problem is I'm not sure how to do this. Maybe we can construct an explicit isomorphism of rings, but I feel like it would be nicer to use some universal property of localization stuff. Is there an easy/clear way to do this?
This is exactly what we want when $\mathscr{F}$ is the structure sheaf, and should imply the result in general. I think $i^* \mathscr{F}$ will correspond to $M \otimes_A A_\mathfrak{q}/\mathfrak{q} A_\mathfrak{q} = M_\mathfrak{q}/\mathfrak{q} M_\mathfrak{q}$, and then taking the stalk at $x$ with be like tensoring with $A_\mathfrak{p}/\mathfrak{q} A_\mathfrak{p}$ over $A_\mathfrak{q}/\mathfrak{q} A_\mathfrak{q}$ so we get $M_\mathfrak{p}/\mathfrak{q} M_\mathfrak{p}$.
For any map of locally ringed spaces $f:X\to Y$ and sheaf of $\mathcal{O}_Y$-modules $\mathcal{F}$, the definition of $f^*\mathcal{F}$ is $f^{-1}\mathcal{G} \otimes_{f^{-1}\mathcal{O}_Y} \mathcal{O}_X$. We can take stalks of both sides to see that $(f^*\mathcal{G})_x = (f^{-1}\mathcal{G})_x\otimes_{(f^{-1}\mathcal{O}_Y)_x}\mathcal{O}_{X,x}$, and use the identity $(f^{-1}\mathcal{G})_x \cong \mathcal{G}_{f(x)}$ to see that this stalk is isomorphic to $\mathcal{G}_{f(x)}\otimes_{\mathcal{O}_{Y,f(x)}}\mathcal{O}_{X,x}$.
We apply the above paragraph as follows: our map is $i:X_y\to X$ and our sheaf $\mathcal{F}$ is on $X$. Substituting in for the last formula about stalks, we get that the stalk we're interested in is isomorphic to $\mathcal{F}_x\otimes_{\mathcal{O}_{X,x}} \mathcal{O}_{X_y,x}$. Writing $\mathcal{O}_{X_y}=\mathcal{O}_X/\mathfrak{m}_y$ and taking stalks, we see that $\mathcal{O}_{X_y,x}=\mathcal{O}_{X_y,x}/\mathfrak{m}_y\mathcal{O}_{X,x}$. Using the identity $M\otimes_{R} R/I=M/IM$, we see that as $\mathcal{F}_x\otimes_{\mathcal{O}_{X,x}} \mathcal{O}_{X_y,x}=\mathcal{F}_x\otimes_{\mathcal{O}_{X,x}} \mathcal{O}_{X_y,x}/\mathfrak{m}_y\mathcal{O}_{X,x}$, we have $(i^*\mathcal{F})_x \cong \mathcal{F}_x/\mathfrak{m}_y\mathcal{F}_x$.
Edit: Let's explain a little more about why $\mathcal{O}_{X_y}=\mathcal{O}_{X}/\mathfrak{m}_y$. We have that $X_y$ is the fiber product of the maps $i_y:\{y\}\to Y$ and $f:X\to Y$, where we assume that the first map is a closed immersion of a closed point because of the OP's use of $\mathfrak{m}_y$ to represent the maximal ideal cutting out $y$ (this is important! if $y$ is not a closed point then $X_y\subset X$ is not a closed subscheme and it is not cut out by a quasicoherent sheaf of ideals). This means that $X_y\to X$ will also be a closed immersion because closed immersions are stable under fiber products. It remains to determine the sheaf of ideals $\mathcal{I}\subset\mathcal{O}_X$ cutting out $X_y$. Pick open affines $\operatorname{Spec} B\subset Y$ and $\operatorname{Spec} A\subset X$ so that $f(\operatorname{Spec} A)\subset \operatorname{Spec} B$. Then as $A,B$ vary, $X_y$ is covered by schemes of the form $\operatorname{Spec} (A\otimes_B B/\mathfrak{m}_y)$ where $\mathfrak{m}_y$ is the maximal ideal cutting out $y$ inside $B$. But by the fact that $M\otimes_R R/I = M/IM$ we see again that these schemes are of the form $A/\mathfrak{m}_yA$. So it makes sense to say $\mathcal{O}_{X_y} = \mathcal{O}_X/\mathfrak{m}_y\mathcal{O}_X$ where when we say $\mathfrak{m}_y$ here we mean the pullback to $\mathcal{O}_X$ of the sheaf of ideals on $Y$ cutting out $y$.