Let $i: \partial X \hookrightarrow X$ be a smooth manifold with boundary and $E \rightarrow X$ an orientable vector bundle. Can any differential form representing the Thom class for $i^*E \rightarrow \partial X$ be extended to a form representing the Thom class for $E \rightarrow X$?

In this MO answer it is explained how to obtain a representative of the Thom class (call this a Thom form) which pulls back to the natural Euler-form given by Chern-Weil theory. This form depends on the choice of: a metric and a compatible connection on $E$, and a cutoff function $\rho:[0,\infty)\rightarrow [0,\infty)$ which is equal to $−1$ near $0$ and equal to $0$ on $[1,\infty)$.

If we are given a Thom form for $i^*E \rightarrow \partial X$ constructed via the above method, one can extend the metric and connection to $E \rightarrow X$ using a partition of unity argument, and use the above method to obtain the required Thom form for $E \rightarrow X$. However, it is not clear to me whether every choice of Thom form can be constructed using this method.

If this general question proves too hard, I am trying to use it to obtain an answer to the following more specific situation: suppose that $i^*E = E_1 \oplus E_2 \rightarrow \partial X$, then one can choose metrics and compatible connections $g_i$ and $\nabla_i$ for $E_i$, and a function $\rho$ as before, and use this to obtain Thom forms $\tau_i \in \Omega^{rk(E_i)}_{cv}(E_i)$, then $\tau_1 \wedge \tau_2$ is a Thom form for $i^*E \rightarrow \partial X$. Can this form be extended to a Thom form for $E \rightarrow X$. And if so, are there choices of metric and connection which give rise to such an extension? One might hope to prove this directly from the formula for the Thom form, but it is not clear to me what happens in the construction of the global angular form of a direct sum.

What I have tried: if X = [0,1], the Poincare Lemma for compactly supported forms on $\mathbb{R}^n$ shows that the result holds for any bundle over $X$. However, already for the next simplest case of $X = D^2$ the question seems non-trivial.