In chapter 2 of Bott & Tu's book, they supposedly prove the Thom isomorphism in the following way. Firstly, by a trivial spectral sequence, they establish the following isomorphism: $$H_{cv}^*(E)\simeq \check{H}^{*-n}\left(\mathfrak{U},\mathcal{H}_{cv}^n\right) $$ between the cohomology with compact vertical support of an arbitrary vector bundle $E$ of rank $n$ and the Cech cohomology of a good cover $\mathfrak{U}$ with values in the cohomology sheaf $\mathcal{H}_{cv}^n:U\mapsto H_{cv}^n(U)$.
Then, in the case of orientable bundles, they construct the Thom class $\Phi\in H_{cv}^n(E)$, whose restriction to each fibre generates its' compact cohomology. This is all fine.
What I don't understand is why this establishes an isomorphism $H_{cv}^n(E)\rightarrow\mathbb{R}$, $\Phi\mapsto 1$ (according to the bottom of page 131). This would surely imply, together with de Rham's theorem, the simplified version of the Thom isomorphism: $$H_{cv}^*(E)\simeq\check{H}^{*-n}\left(\mathfrak{U},\mathcal{H}_{cv}^n\right)\simeq \check{H}^{*-n}\left(\mathfrak{U},\underline{\mathbb{R}}\right)\simeq H_{dR}^{*-n}(M) $$
but I cannot find a proper justification for it. The obvious answer would be the Leray-Hirsch theorem. However, the problem with applying L-H here is that it would render the entire method of proof explained above useless - it directly yields: $$H_{cv}^*(E)\simeq H_{dR}^{*-n}(M) $$ not just the isomorphism $H_{cv}^n(E)\rightarrow\mathbb{R}$.
Could this have been the intention of the authors or am I missing something obvious? Or is this something that has been overlooked, being a rather outdated subject?
Thank you in advance.