This is true without any further assumptions on $M,V,W$ and discussed quite thoroughly in chapter 7 of Conlon's wonderful book "Differentiable Manifolds". The proof given in the text is quite elegant so I can't resist recalling the argument.
There is an obvious $C^{\infty}(M)$-bi-linear map $\alpha \colon \Gamma(V) \times \Gamma(W) \rightarrow \Gamma(V \otimes W)$ given by $\alpha(\xi,\eta)_p = \xi_p \otimes \eta_p$ which gives rise to a map $\alpha \colon \Gamma(V) \otimes_{C^{\infty}(M)} \Gamma(W) \rightarrow \Gamma(V \otimes W)$. We want to show that $\alpha$ is an isomorphism.
When $V$ and $W$ are trivial bundles of rank $n$ and $k$ respectively, one can choose $n$ pointwise linearly independent sections $\xi_i \in \Gamma(V)$ and $k$ pointwise linearly independent sections $\eta_j \in \Gamma(W)$ and then $\Gamma(V \otimes W)$ is easily seen to be a free $C^{\infty}(M)$ module with basis $\{ \xi_i \otimes \eta_j \}_{i,j}$ and so $\alpha$ is an isomorphism of $C^{\infty}(M)$ modules.
If $V$ and $W$ are not trivial, we interpret them as a direct summands of some trivial bundles. That is, we find vector bundles $V^{\perp}$ and $W^{\perp}$ such that $V \oplus V^{\perp}$ and $W \oplus W^{\perp}$ are trivial bundles. Then, by looking at the diagram
$$ \require{AMScd}
\begin{CD}
\Gamma((V \oplus V^{\perp}) \otimes (W \oplus W^{\perp})) @<{\tilde{\alpha}}<< \Gamma(V \oplus V^{\perp}) \otimes_{C^{\infty}(M)} \Gamma(W \oplus W^{\perp}) \\
@AAA @AAA \\
\Gamma(V \otimes W) @<{\alpha}<< \Gamma(V) \otimes_{C^{\infty}(M)} \Gamma(W)
\end{CD} $$
we see that $\tilde{\alpha}$ is injective by the previous paragraph, and we can check directly that both vertical arrows (that are defined with the help of the injective bundle maps $v \mapsto (v,0)$ of $V \rightarrow V \oplus V^{\perp}$ and $w \mapsto (w,0)$ of $W \rightarrow W \oplus W^{\perp}$) are injective, showing that $\alpha$ is injective. Similarly, by reversing the direction of the vertical arrows and replacing them with projections, we see that $\alpha$ is also surjective.
The crux of the proof is the result that every bundle $V$ over $M$ can be realized as a subbundle of a trivial bundle $F$ (stated in terms of the module of global sections, this means that $\Gamma(V)$ is a $C^{\infty}(M)$-projective module). This can be shown by constructing an epimorphism $\psi \colon F \rightarrow V$ of vector bundles from a trivial bundle $F$ onto $V$ and using a fiber metric to split $F$ as an inner direct sum $F = \ker(\psi) \oplus \ker(\psi)^{\perp}$ with $\ker(\psi)^{\perp} \cong V$.
The construction of $\psi$ uses partition of unity. When $M$ is compact, one takes a partition of unity $\{\lambda_i\}_{i=1}^n$ subordinate to a cover $\{U_i\}_{i=1}^n$ of $M$ over which $V$ trivializes with generating global sections $\xi_i^j \in \Gamma(U_i,E|_{U_i})$. Then, one can define global sections $\sigma_i^j = \lambda_i \xi_i^j$ by zero extension outside $U_i$. We obtain finitely many sections, and by taking the trivial bundle with the vector space $X = \mathrm{span} \{ \sigma_i^j \} \subseteq \Gamma(V)$ as fiber, and defining $\psi \colon M \times X \rightarrow V$ as $\psi(p,\sum \sigma_i^j) = \sigma_i^j(p)$ we obtain the required map.
Addendum:
- One can use the ideas described in the proof above to show Swan's theorem about the equivalence of categories between the category of (smooth, finite rank) vector bundles over $M$ and the category of projective finitely generated $C^{\infty}(M)$ modules.
- In the algebraic / holomorphic setting, this fails badly, at least for projective varieties. While in the smooth category, you can identify a vector bundle with the module of global sections, in other settings the module of global sections doesn't hold enough information about the vector bundle and one should consider the sheaf of sections instead. This failure also provides a geometric example of why one needs to sheafify when taking tensor products (one expects that the sheaf of sections of the tensor product will be the tensor product of the sheaves of sections and so $\Gamma(\mathcal{O}(-1) \otimes \mathcal{O}(1)) \cong \Gamma(E(1) \otimes E(-1)) \cong \Gamma(E(0)) = \mathbb{C}$ which obviously cannot hold if you don't sheafify).
- The fact that every (continuous) finite rank vector bundle is a direct summand of a trivial bundle is true even if we merely assume that $M$ is compact and Hausdorff. One can get rid of the assumption that $M$ is compact when $M$ is a manifold, but cannot get rid of it in general. In Hatcher's "Vector Bundles and K-Theory", it is shown that the tautological line bundle over $M = \mathbb{RP}^{\infty}$ is not a direct summand of a trivial bundle using characteristic classes.
No, this is not true. Obviously it is necessary for the fibers to all fit into short exact sequences, but it is not sufficient.
For a counterexample, let $X$ be any space that has a nontrivial vector bundle $E$ of rank $n$. Let $F$ denote the trivial vector bundle over $X$ of rank $n$, and let $0$ denote the vector bundle of rank $0$ over $X$. Then for each $x\in X$, there is a short exact sequence $0\to 0_x\to E_x\to F_x\to 0$ of fibers at $x$, since $E_x$ and $F_x$ are both $n$-dimensional vector spaces and $0_x$ is the trivial vector space. But there is no short exact sequence of vector bundles $0\to 0\to E\to F\to 0$, since this would imply $E\cong F$, but $E$ is nontrivial.
However, as Mariano Suárez-Alvarez commented, if you already have maps of vector bundles $E\to F$ and $F\to G$, then it is true that the sequence $0\to E\to F\to G\to 0$ is exact iff $0\to E_x\to F_x\to G_x\to 0$ is exact for each $x\in X$. I can't prove this without knowing what your definition of "exact sequence of vector bundles" is (indeed, one such definition is that a sequence is exact iff it is exact on each fiber).
Best Answer
Yes. Every short exact sequence of vector bundles splits; for example, if we pick a Riemannian metric on the bundles we can split them by taking an orthogonal complement. So WLOG every short exact sequence has the form
$$0 \to P \to P \oplus R \to R \to 0$$
and then we get a split exact sequence of spaces of sections
$$0 \to \Gamma(M, P) \to \Gamma(M, P) \oplus \Gamma(M, R) \to \Gamma(M, R) \to 0.$$