Let $E$ be a real oriented vector bundle of rank $k$ over a differentiable manifold $M$ of dimension $n$. By ''orientable'', I mean that the structure group of the bundle can be reduced to $GL^+(k,\mathbb{R})$.
We have (since $E$ is orientable) a Thom class $[\Phi] \in H^k_{cv}(E)$ ($*$) which restricts to a generator of $H^k_c(F) \cong H^k_c(\mathbb{R}^k) \cong \mathbb{R}$ for every fiber $F$, since $\int_F i^*(\Phi) = 1$ for every fiber $F$, where $i:F \to E$ is the inclusion.
I'd like to know if the existence of such a cohomology class guarantees that $E$ is orientable. More precisely, I'm wondering if the following holds:
Claim: Let $E$ be a real vector bundle of rank $k$ over $M$. There exists a closed form $\Phi \in \Omega^k_{cv}(E)$ such that $\int_F i^*(\Phi) = 1$ for every fiber $F$ (where $i: F \to E$ is the inclusion) if and only if $E$ is orientable.
If the claim held, then we could define an orientation of a vector bundle as a choice of such a form $\Phi$.
The reason (I may be wrong, of course) I think it should hold is that $\int_F i^*(\Phi) = 1$ implies $[i^*\Phi]$ is a generator of $H^k_c(F)$, and gives a ''sort of'' (can this be formalized?) orientation of $F$ for every fiber $F$. Since $\Phi$ is defined globally and is continuous, this choice of orientation is consistent, and then defines an orientation for $E$.
$(*)$ $\Phi \in \Omega^k_{cv}(E)$ is a closed differential form with compact vertical support. See for example Bott-Tu's Differential forms in Algebraic Topology.