Suppose we have two affine open subschemes $X=Spec(A)$ and $Y=Spec(B)$ of a separated scheme $Z$ (let's even assume that $Z$ is a $k$-variety). Then we know that the intersection $X\cap Y$ is an affine subscheme.
My question is: what is the coordinate ring of $X\cap Y$?
The way to show that the intersection is affine is to notice that $X\cap Y\cong (X\times_k Y)\cap \Delta$, where $\Delta$ is the diagonal. Since $Z$ is separated, $(X\times_k Y)\cap \Delta$ is a closed subscheme of the affine scheme $X\times_k Y=Spec(A\otimes_k B)$ and hence must be affine.
So now the question becomes: what is the surjective ring map inducing this closed immersion? I want to say it's just $f\colon A\otimes_k B\rightarrow k[A,B], \, a\otimes b\mapsto ab$, where by $k[A,B]$ I just mean the $k$-algebra generated by $A$ and $B$. Then we would just have $X\cap Y=Spec((A\otimes_k B)/I)$, where $I=ker(f)$.
Ravi Vakil uses a similar map in Proposition 10.1.3 to show that the diagonal morphism is a locally closed embedding. But I don't understand what happens when we restrict to $X\times_k Y$.