Vector bundles with connections over the same manifold $M$ make up a category. Indeed, let $E, E' \twoheadrightarrow M$ be vector bundles with connections $\nabla$ and $\nabla'$. A morphism between vector bundle with connections is a vector bundle morphism $F: E \to E'$ such that for all sections $s \in \Gamma (E)$ it holds $F(\nabla s)=\nabla' F(s)$.
Does there exist a similar category for vector bundles over arbitrary (non-fixed) manifolds? I was hoping that the above construction could be generalized in such a way that the natural morphism on the pull-back vector bundle induces a morphism on the pull-back vector bundle with the pull-back connection, but I am pretty sure the answer is no, as there is no sensible way to pull-back a section through a generic bundle morphism.
Let's try to define the category $\mathcal C$ of contravariant vector bundles equipped with a connexion, I haven't checked the details and I don't know if it is the right way of seeing it, so be careful. An object in $\mathcal C$ is a triple $(M,E,\nabla)$ with $M$ a manifold, $E$ a vector bundle on it and $\nabla$ a connexion on $E$, that is, a bundle morphism $E\to E\otimes T^*M$ satisfying the Leibniz rule. If $M'$ is another manifold and $f\colon M'\to M$ is a smooth map, there is a functorial morphism $f^*T^*M\to T^*M'$, the pullback of forms. Call $f^*\nabla$ the composition $$f^*\nabla\colon f^*E\to f^*(E\otimes T^*M)\cong f^*E\otimes f^*T^*M\to f^*E \otimes T^*M'$$ it is a connection on $f^*E$.
Now, if $(M',E',\nabla')$ is another object in $\mathcal C$, a morphism $(M',E',\nabla')\to (M,E,\nabla)$ consists of a smooth map $f\colon M'\to M$ and a vector bundle morphism $F: f^*E\to E'$ such that
commutes. I guess you can also define the category of covariant vector bundles equipped with a connexion by taking the bundle morphism in the opposite direction, $F: E'\to f^*E$.