Let $\pi: E \rightarrow M$ be a vector bundle and let $f: N \rightarrow M $ be a continuous map. Let $f^{*} E$ be pullback bundle
Let $\Gamma(E)$ be the module of section of the vector bundle $E$ and $\Gamma(f^*E)$ the module of section for the pullback bundle.
Now in this book Elements of Noncommutative Geometry at page 62 they have
Proposition 2.12. There is an isomorphism of $C(N)$-modules, $$ \Gamma\left(N, f^{*} E\right) \simeq \Gamma(M, E) \otimes_{C(M)} C(N) $$Proof. We proceed by interpreting geometrically the right hand side. We can regard simple tensors like $s \otimes g$ as functions $N \rightarrow E: y \mapsto s(f(y)) g(y)$ of a particular type, to wit, as sections of $E \rightarrow M$ along the map $f: N \rightarrow M$. A section along a map $f: N \rightarrow M$ is defined as a continuous map $\sigma: N \rightarrow E$ such that $\pi \circ \sigma=f$. For instance, a vector field $X$ over $N$ defines a vector field along $f$, namely $T f \circ X$. Now, there is a one-to-one-correspondence, in fact an isomorphism of $C(N)$-modules, between sections $\sigma$ of $E \rightarrow M$ along the map $f: N \rightarrow M$ and sections of the pullback bundle $f^{*} E \rightarrow N$ given by $y \mapsto \tilde{f}^{-1} \sigma(y)$; recall that $\tilde{f}$ is an isomorphism on each fibre.
I am not understanding the proof. How the isomorphism of the fibers implies that the theorem is true?