If $A$ is a commutative ring, $M\in A\text{-mod}$ and $\mathfrak{p}$ is a prime ideal of $A$ then it is an easy fact from commutative algebra that $M\otimes_{A}A_{\mathfrak{p}}\cong M_{\mathfrak{p}}$.
Let $X$ be a scheme and $\mathcal{F}$ be a coherent sheaf on $X$. For $U$ open in $X$ and $x\in U$, is it true that $$\Gamma(U,\mathcal{F})\otimes_{\Gamma(U,\mathcal{O}_X)}\mathcal{O}_{X,x}\cong\mathcal{F}_{x}?$$
Rough sketch. Is it true for affine $X$ ? Yes, results from what you recalled. Now for a general $X$, take an open affine neighbourhood of $x$ in $X$. How can you rewrite the two sides of the desired isomorphism relation ?
Proof. To prove $\Gamma(U,\mathcal{F})\otimes_{\Gamma(U,\mathcal{O}_X)}\mathcal{O}_{X,x}\cong\mathcal{F}_{x}$ for any open $U$ open in $X$, as the question is (eminently !) local on $X$, if suffices to prove $\Gamma(U,\mathcal{F}|_U)\otimes_{\Gamma(U,\mathcal{O}_X|_U)}({{\mathcal{O}_X|}_U})_x \cong \mathcal{F}_{x}$. This shows that one can assume that $U = X$, so that we have to show $\Gamma(X,\mathcal{F})\otimes_{\Gamma(X,\mathcal{O}_X)}\mathcal{O}_{X,x}\cong\mathcal{F}_{x}$. In light of the previous argument, that is, by an analogue argument, by taking an open affine neighbourhood $U = \textrm{Spec}(A)$ of $x$ in $X$ and restricting, on can assume that $X =\textrm{Spec}(A)$ is affine, in which case the assertion results from $M\otimes_{A}A_{\mathfrak{p}}\cong M_{\mathfrak{p}}$ with $M = \Gamma(X,\mathcal{F})$.
Bonus question : where did I use the coherence of $\mathcal{F}$ as $\mathcal{O}_X$-module ? ;-)
Hint : what is the essential image of the functor $M\mapsto \widetilde{M}$ from $\mathfrak{Mod}_{/A}$ to $\mathfrak{mod}_{/ \mathcal{O}_X}$ ?