This question regards a part of the proof of the so called *surgery step*, in Wall's [book][1] "surgery on compact manifolds", Theorem 1.1.

**Setting**

$M^m$ smooth manifold, $X$ CW complex,  $\phi :M\to X$ continuous map, $\nu\to X$ a rank-$v$ vector bundle and $F:TM\oplus \phi^*\nu \oplus  \varepsilon^q \to \varepsilon^{m+q+v}$ is a given stable trivialization of $TM\oplus \phi^*\nu$.



The second part of Theorem 1.1. asserts that if $f:\mathbb S^r\times \mathbb D^{m-r}\to M$ is an *embedding*, $m\geq r+2$ and $f_0=f|_{\mathbb S^r\times\{0\}}$ makes this diagram commute:

$\require{AMScd}$
\begin{CD}
    \mathbb S^r @>>f_0>  M\\
    @VViV            @VV{\phi}V \\
    \mathbb D^{r+1} @>>Q> X  \\
\end{CD}
for some $Q$, *then we can perform the surgery step*, i.e. denoting by $W^{m+1}$ the trace of the surgery along $f$, $\phi$ extends to $W$, yielding $ \phi_W: W\to X$, and also the trivialization $F$ extends to a stable trivialization of $TW\oplus \phi_W^*\nu$.
In other words, we obtain a cobordism of the normal maps.

**Question**
The proof of this fact takes a few lines in the book (pg. 11, 3rd and 4th paragraph) and 
relies on the fact that $TW\oplus \phi_W^*\nu$ restricted to the handle is trivial (the handle is contractible) and that this trivialization coincides with that induced by $F$ so that the two glue to a trivialization over $W$.
> Why the two trivialization coincide?
I don't see any good reason why the obstruction to extend the trivialization over the core of the handle should vanish in general.

  [1]: https://www.maths.ed.ac.uk/~v1ranick/books/scm.pdf