The stalk of a tensor product sheaf is the tensor product of the stalks

By [F1] and [L1], it is enough to identify the stalk of the tensor-product presheaf $U⟼\mathcal{F}(U){\otimes }_{{\mathcal{O}}_X(U)}\mathcal{G}(U)\mathrm{.}$

For every neighbourhood $U$ of $x$, the germ maps $\mathcal{F}(U)\to {\mathcal{F}}_x$, $\mathcal{G}(U)\to {\mathcal{G}}_x$, and ${\mathcal{O}}_X(U)\to {\mathcal{O}}_{X,x}$ induce an ${\mathcal{O}}_X(U)$-balanced pairing into ${\mathcal{F}}_x{\otimes }_{{\mathcal{O}}_{X,x}}{\mathcal{G}}_x$. Hence there is a homomorphism $\mathcal{F}(U){\otimes }_{{\mathcal{O}}_X(U)}\mathcal{G}(U)⟶{\mathcal{F}}_x{\otimes }_{{\mathcal{O}}_{X,x}}{\mathcal{G}}_x,$ and these maps are compatible with restriction. Therefore they induce a map $\mathrm{\Phi }:(\mathcal{F}{\otimes }_{p,{\mathcal{O}}_X}\mathcal{G}{)}_x⟶{\mathcal{F}}_x{\otimes }_{{\mathcal{O}}_{X,x}}{\mathcal{G}}_x\mathrm{.}$

Conversely, if $s_x\in {\mathcal{F}}_x$ and $t_x\in {\mathcal{G}}_x$ are represented by sections $s\in \mathcal{F}(U)$ and $t\in \mathcal{G}(U)$ on the same neighbourhood $U$ of $x$, define $\mathrm{\Psi }(s_x\otimes t_x):=(s\otimes t{)}_x\mathrm{.}$ If the representatives are changed on a smaller neighbourhood, the represented germ of $s\otimes t$ is unchanged there, so $\mathrm{\Psi }$ is well defined on simple tensors and extends linearly to a homomorphism $\mathrm{\Psi }:{\mathcal{F}}_x{\otimes }_{{\mathcal{O}}_{X,x}}{\mathcal{G}}_x⟶(\mathcal{F}{\otimes }_{p,{\mathcal{O}}_X}\mathcal{G}{)}_x\mathrm{.}$

On a simple tensor represented on one neighbourhood, $\mathrm{\Phi }$ and $\mathrm{\Psi }$ plainly undo each other. Since both sides are generated by simple tensors, they are inverse isomorphisms. Combining this with step 1.1 gives the stated isomorphism for the sheaf tensor product.