If $\mathscr{H}$ is an $\mathscr{O}_X-$module that is locally of finite presentation, then for every $\mathscr{O}_X-$module $\mathscr{G}$ and every $x\in X$, the natural map $$\mathscr{H}om_{O_X}(\mathscr{H,G})_x\longrightarrow Hom_{O_{x,X}}(\mathscr{H_x,G_x})$$ is an isomorphism.
This is quite easy to prove when $\mathscr{G}$ is quasi-coherent by using the fact that $$\mathscr{H}om_{O_X}(\tilde{M},\tilde{N})\cong\widetilde{Hom_R(M,N)}$$ when $M$ is of finite presentation.
But I can't get further.
First prove it when $\mathscr{H}=\mathcal{O}_X^n$ is a free $\mathcal{O}_X$-module : in that case $\mathscr{H}om_{\mathcal{O}_X}(\mathscr{H},\mathscr{G})\simeq\mathscr{G}^n$. The result follows easily : $\mathscr{H}om_{\mathcal{O}_X}(\mathscr{H},\mathscr{G})_x\simeq\mathscr{G}^n_x$ and $\mathscr{H}om_{\mathcal{O}_{X,x}}(\mathscr{H}_x,\mathscr{G}_x)=\mathscr{H}om_{\mathcal{O}_{X,x}}(\mathscr{O}_{X,x}^n,\mathscr{G}_x)\simeq\mathscr{G}_x^n$. (The morphisms are the natural ones).
Now, for any $x\in X$, find a neighborhood $U$ such that $\mathscr{H}$ admit a free presentation. You have then $$ \require{AMScd} \begin{CD} \mathscr{H}om_{\mathcal{O}_U}(\mathcal{O}_U^m,\mathscr{G}|_U)_x@>>>\mathscr{H}om_{\mathcal{O}_U}(\mathcal{O}_U^n,\mathscr{G}|_U)_x@>>>\mathscr{H}om_{\mathcal{O}_U}(\mathscr{H},\mathscr{G}|_U)_x@>>>0\\ @VVV@VVV@VVV\\ \mathscr{H}om_{\mathcal{O}_{X,x}}(\mathcal{O}^m_{X,x},\mathscr{G}_x)@>>>\mathscr{H}om_{\mathcal{O}_{X,x}}(\mathcal{O}^n_{X,x},\mathscr{G}_x)@>>>\mathscr{H}om_{\mathcal{O}_{X,x}}(\mathscr{H}_x,\mathscr{G}_x)@>>>0 \end{CD} $$ By the result in the free case, the first two vertical arrows are isomorphisms, hence by diagram chase, the last one is also an isomorphism.