Prove a ring isomorphism of stalks

I will put my solution.

First as you said, everything is local so we can assume $X=\text{Spec}(B)$ and $Y=\text{Spec}(A)$ and we change the notation to $x=\mathfrak{q}$, $y=\mathfrak{p}$.

Now as $\text{Spec}(k(y))=\text{Spec }A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}$ and $X\times_Y \text{Spec}(k(y))=\text{Spec}(B\otimes_A A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p})$ we have the cartesian diagram

$$\require{AMScd} \begin{CD} \text{Spec}(B\otimes_A A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}) @>>> \text{Spec}(A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}) \\ @VVV @VVV\\ \text{Spec}(B) @>>> \text{Spec}(A) \end{CD}$$

But now as $\text{Spec}(A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p})\rightarrow \text{Spec}(A)$ is a universal topological embedding (see for example Görtz and Wedhorn Remark 4.21) we have that $\text{Spec}(B\otimes_A A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p})\rightarrow \text{Spec}(B)$ is a homeomorphism onto $X_y$, in particular $x\in X_y$ correspond to the unique $\tilde{\mathfrak{q}}$ such that its preimage is $\mathfrak{q}$ under $B\rightarrow B\otimes_A A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}$. Hence we have $$\mathcal{O}_{X_y,x}=(B\otimes_A A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p})_\tilde{\mathfrak{q}}.$$

Finally as the multiplicative system of $\tilde{\mathfrak{q}}$ in $B\otimes_A A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}$ is equal to the multiplicative system induced by localization at $\mathfrak{q}$ as a $B$-algebra we have:

$$\begin{align}\mathcal{O}_{X_y,x} =& (B\otimes_A A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p})_\tilde{\mathfrak{q}} \\ =& ( A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}\otimes_A B)\otimes_B B_\mathfrak{q} \\ =& A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}\otimes_A B_\mathfrak{q} \\ =& A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}\otimes_{A_\mathfrak{p}} B_\mathfrak{q}\\ =& B_\mathfrak{q}/\mathfrak{p}B_\mathfrak{q} \\ =& \mathcal{O}_{X, x}/\mathfrak{m}y \, \mathcal{O}_{X,x} \end{align}$$

The Stacks Project might not be a great place to learn algebraic geometry for the first time, but you might look here or just google fiber product of schemes (or just of locally ringed spaces) to learn more about the definitions (it's kind of hairy, just to warn you). It's definitely good to know the universal property, but I think at the end of the day you'll be happier to just know an explicit construction.

It turns out that there's an open cover of the scheme-theoretic fiber $X\times_Y \mbox{Spec}(\kappa(y))$ that's easy-ish to describe (see Lemma 25.17.4 in the link above). In our case, it's covered by affine opens like the one you described, $\mbox{Spec}(B\otimes_A \kappa(y))$. First, let's make the identification $B\otimes_A \kappa(y) \cong B \otimes_A A/\mathfrak p \otimes_A A_\mathfrak{p} \cong (B/\mathfrak p B)_\mathfrak{p}$, where that last $\mathfrak p$ in the subscript means to invert the image of $A-\mathfrak p$ in $B/\mathfrak pB$ (if this tensor product translation is foreign, see Atiyah-MacDonald chapters 2 & 3 or google around for quotient and localization as tensor products). Theres a point here corresponding to $x$, which is the prime ideal of $(B/\mathfrak p B)_\mathfrak{p}$ corresponding to $\mathfrak q$ under the prime ideal correspondences (prime ideals in the quotient, prime ideals in the localization) for the maps $B \to B/\mathfrak p B \to (B/\mathfrak p B)_\mathfrak{p}$. If you trace all of these isomorphisms and correspondences through, you'll see that the stalk of $X\times_Y \kappa(y)$ at that point corresponding to $x$ is isomorphic to $((B/\mathfrak p B)_\mathfrak{p})_\mathfrak{q}) \cong B_\mathfrak{q}/\mathfrak{p}B_\mathfrak{q}$, as desired (that's the LHS).

Sorry I haven't traced all of that myself here -- anyway, it's probably more useful for you to do that.