The only natural vector bundles I know exist on all smooth manifolds are jet-spaces and everything they generate; you can apply the operations that are natural on vector spaces such as tensor, wedge product, dual, etc. From technical considerations I will actually restrict myself to cotangent bundles and cotangent jet bundles; because they pullback as all vector bundles do.
I wonder if we can prove those are everything; that is, if we consider the category $C$ of smooth manifolds, and construct a vector bundle on each one of them so that all pullbacks are compatible, must the vector bundles be those combinations of conatural vector bundles?
I.e if have the category of $(M, V)$ with $V$ a bundle on $M$, with the projection to the category of manifolds, then any section (functorial one that flips arrows) assigns each manifolds one of the standard bundles we talked about (generated by cotangent).
This is false. As an example, consider the circle $S^1$. All of the jet spaces of $C^\infty(S^1,\mathbb{R})$ are trivial, and since smooth functors on $\mathsf{Vect}_{\mathbb{R}}$ preserves identity morphisms, any bundle arising from these functors must also be trivial. This means that nontrivial bundles such as the Möbius strip cannot arise this way.
Your question about a "section" of the forgetful functor $F:\mathsf{SVB}_{\mathbb{R}}\to\mathsf{Diff}$ doesn't seem to make sense (perhaps because it doesn't take morhpisms into account). $F$ is a covariant functor, so any "section" of $F$ must also be covariant. Such functors certainly exist (e.g. define $G(M)=M\times\mathbb{R}$ and $G(f)=(f,f\times\operatorname{id}_{\mathbb{R}})$), but they don't do what you seem to want them to.