Why any short exact sequence of vector spaces may be seen as a direct sum?

In general you cannot see $V$ as a direct sum $U \oplus W$ when these are modules over a general ring $R$. For example consider the short exact sequence: $$\newcommand\Z{\mathbb{Z}}0 \to \Z/2\Z \to \Z/4\Z \to \Z/2\Z \to 0,$$ it is not true that $\Z/4\Z$ is isomorphic to $\Z/2\Z \oplus \Z/2\Z$.

However when you are working with vector spaces over a field (and since you tagged this "differential topology" I assume you're actually interested in de Rham cohomology, which is over $\mathbb{R}$), then yes, it follows that $V$ is isomorphic to $U \oplus W$. A simple way to see that in finite dimension is through the rank-nullity theorem (and the fact that two vector spaces are isomorphic iff they have the same dimension).

In general though, there is still an isomorphism, even if $U$, $V$, $W$ have infinite dimension (and you assume the axiom of choice I guess). Every vector space is free, ie. has a basis. Choose a basis $\{w_i\}$ of $W$. Since $B$ is surjective, there are elements $v_i \in V$ such that $B v_i = w_i$. Define a linear map $C : W \to V$ by $Cw_i = v_i$. Since $\{w_i\}$ is a basis this determines a (unique) linear map $C$. Then it is a simple exercise to show that the following map is an isomorphism: $$\begin{align} U \oplus W & \to V\\ (u, w) & \mapsto (Au, Cw) \end{align}$$

The conclusion is false in general. Pick cyclic groups ${\mathbb Z}_p$, ${\mathbb Z}_q$, and ${\mathbb Z}_{pq}$. You can easily cook up a short exact sequence $$ 0\to{\mathbb Z}_p\to{\mathbb Z}_{pq}\to{\mathbb Z}_q\to0, $$ but the group in the center is not the direct product of the groups on its sides.

The conclusion holds if, e.g., you have a short exact sequence of vector spaces.