I need some help in finding the solution to the following problem:
Let $E$, $F$ be vector bundles (of finite rank, if needed) over a manifold $M$. Consider the set $\operatorname{Hom}(E,F)$ of all bundle morphisms from $E$ to $F$. Then, $\operatorname{Hom}(E,F)$ is isomorphic to $E^*\otimes F$.
I believe that the first step I took in my attempt to solve this is correct: fixing $x\in M$, I have $E_x$, $F_x$ vector spaces. Then, I can consider the map $$ E^*_x\otimes F_x \rightarrow \operatorname{Hom}(E_x,F_x) $$ given by $$ f(x)\otimes v(x) \rightarrow [w(x)\mapsto f(v)w(v)]. $$
This shows the result on fibers. I do not know how to extend its validity to the entire bundle. Can someone give me any suggestion?
Define a map $\Phi:E^*\otimes F\to \mathrm{Hom}(E,F)$ by your map on each fiber. You have shown this is a linear isomorphism on each fiber, so it suffices to show it is a continuous map on the total spaces. This statement is local on $M$, so we may assume $E\cong M\times \mathbb{R}^m$ and $F\cong M\times\mathbb{R}^m$ are trivial. But then $\Phi:M\times((\mathbb{R}^m)^*\otimes\mathbb{R}^n)\to M\times (\mathrm{Hom}(\mathbb{R}^m,\mathbb{R}^n))$ is just given by $\Phi(m,x)=(m,\varphi(x))$ for some particular linear isomorphism $\varphi:(\mathbb{R}^m)^*\otimes\mathbb{R}^n\to \mathrm{Hom}(\mathbb{R}^m,\mathbb{R}^n)$ (namely, your map $E_x^*\otimes F_x\to \mathrm{Hom}(E_x,F_x)$ in the particular case that $E_x=\mathbb{R}^m$ and $F_x=\mathbb{R}^n$). This $\varphi$ is a continuous map, and so it follows that $\Phi$ is continuous.
(The key point here is that your description of the isomorphism $E_x^*\otimes F_x\to \mathrm{Hom}(E_x,F_x)$ depended only on the vector space structures of $E_x$ and $F_x$, so the same description still holds when we replace $E_x$ and $F_x$ with $\mathbb{R}^m$ and $\mathbb{R}^n$ via any chosen isomorphisms $\mathbb{R}^m\to E_x$ and $\mathbb{R}^n\to F_x$. It follows that in a local trivialization, $\Phi$ is just the same linear isomorphism on every fiber.)