A vector bundle $(E,p, B)$ can be viewed as a disjoint union of all its fibers, $E = \coprod_{x\in B}E_x$. Through vector bundle charts, we can describe $E$, at least in a neighbourhood of the base space, $U\subset B$, as $$p^{-1}(U)\simeq_{\Phi} U \times \mathbb{R}^n ,$$ where the identification is made via a bundle chart $\Phi$.
My question is the following, from a categorical point of view, can we always express a disjoint union, locally, as a cross product? Or is this property, what characterizes vector bundles?
Given an index set $I$. If you have a disjoint union of $I$-many copies of $X$, then this is by definition the set
$$X\times I.$$
So the only question remaining is whether this also yields the same topology in general (and not just for vector bundles). And the answer is Yes, if we equip $I$ with the discrete topoloy:
Proof.
We have two topologies on $X\times I$ which we have to show are equal:
$\Rightarrow:\;$
Assume $U_i:=\{x\in X\mid (x,i)\in U\}$ is open for all $i\in I$. Then $U$ is the union of the sets $U_i\times\{i\}$ and $U_i$ is open in $X$ and $\{i\}$ is open in $I$.
$\Leftarrow:\;$
Assume $U$ is the union of sets $U_{X,k}\times U_{I,k}$ for some index set $K$. Define $\mathfrak U_i:=\{U_{X,k}\mid k\in K,i\in U_{I,k}\}$. Then
$$U_i =\left\{x\in X\mid (x,i)\in U\right\} =\bigcup_{U\in \mathfrak U_i}U. $$
Since all the $U\in\mathfrak U_i$ are open, and $U_i$ is their union, $U_i$ is open too.
$\square$
If however $I$ is not equipped with the discrete topology, then this is not true anymore. A topology is discrete if and only if all points $\{i\}$ are open sets. So if it is not discrete, there is some $i\in I$, so that $\{i\}$ is not open anymore. This means
$$X\times\{i\}$$
is not open in $X\times I$ w.r.t to the product topology. But $X\times\{i\}$ is open w.r.t. the sum topology. So the product $X\times I$ (the simplest bundle) cannot be seen as a disjoint union of $I$-many copies of $X$, at least not using the obvious identity map $\mathrm{id}$ as homeomorphism.