$E$-orientation of a closed manifold induce $E$-orientation of normal bundle: passage in the proof.

Question

I'm trying to follow Kochman's proof of the well-known result

For a ring spectrum $E$, and closed manifold $M^n$ together with an embedding in $\mathbb{R}^{n+k}$, the following are equivalent:

1. the vector bundle is $E$-oriented

2. $M^n$ is $E$-oriented.

In the proof of $1\Rightarrow 2$, he makes use of this commutative diagram

My question is the following: why the capping with $U$ and $U'$ on the left and right column respectively induce an isomorphism?

They much resemble Thom isomorphisms. The problem is that Thom isomorphisms are proved in the following page. I'm wondering if there is a way to show that it is an iso without the full power of Thom Isomorphism theorem, or if it is just a "wrong placing" of this proposition, which should go after Thom Isos theorem.