I can prove the only if part. My attempt to prove if part is the following:
Given $n$ global sections $s_1, s_2, ..., s_n$ of a vector bundle $E$ on a smooth manifold $M$ such that they form a basis at each fibre, we construct a smooth map $M\times \mathbb R^n \to E$ defined as $$(m, (x_1, ..., x_n))\mapsto x_1s_1+...+x_ns_n$$ It is easy to see this map is injective, smooth and is linear isomorphism on fibers. However, I fail to prove the inverse is smooth because of two elementary reasons:
If $s$ is an arbitrary smooth local section of $M$ to $E$, then s has a unique decomposition $s=x_1s_1(m)+...+x_ns_n(m)$ where $x_i$ are local functions. How do we show that they are smooth?
If the statement above is shown, we can show every smooth section from $M$ to $E$ maps to a section from $M$ to $M\times \mathbb R^n$. Can this necessarily imply the map from $M\times \mathbb R^n$ to $E$ is smooth?
Of course, my direction may be wrong, so if the standard approach is more straightforward, please let me know.
The trick is to work with a local trivialization of $E$ since smoothness of a map is a local property. Now, for both parts 1 and 2 use the fact that inverse of an invertible matrix $A$ depends smoothly on its entries and hence, solution vector of a linear system $Ax=b$ depends smoothly on $A$ and $b$.