I put this question in math.stackexchange but since I had no answer
Let $\pi: E \rightarrow B$ be a fiber bundle with abstract fiber $F$ and let $f: B^{\prime} \rightarrow B$ be a continuous map. Define the pullback bundle by $$ f^{*} E=\left\{\left(b^{\prime}, e\right) \in B^{\prime} \times E \mid f\left(b^{\prime}\right)=\pi(e)\right\} \subseteq B^{\prime} \times E $$
Let $\Gamma(E)$ be the module of section of the vector bundle $E$ and $\Gamma(f^*E)$ the module of section for the vector bundle $\pi':f^*E\rightarrow B $
My goal is to proof that $\Gamma(E)$ and $\Gamma(f^*E)$ are isomorphic.
Define the map $\phi:\Gamma(E)\rightarrow \Gamma(f^*E)$ by $\phi(\sigma)(p)= \sigma(f(p))$ for $p\in B'$
This map is surjective since for a section $\sigma* \in \Gamma(f^*E)$ we can define the section $\sigma \in \Gamma(E)$ by $\sigma(x)=\sigma(p)$ where $\{x\in B: f(x)=p, p \in B'\}$
How to proof that this map is injective?
This map is neither injective nor surjective. As a counterexample, as pointed out in the comments, consider an inclusion $\{*\}\rightarrow B$ (which corresponds to the unique point $x\in B$ image of $*$), then the pullback bundle is simply $E_x=p^{-1}(x)$ and the map $\Gamma(E)\rightarrow \Gamma(E_x)\simeq E_x$ is simply "evaluation at $x$". A section of $E_x\rightarrow\{*\}$ is simply an element in the fiber (which is always non-empty) so $\Gamma(f^*E)\simeq E_x\neq\emptyset$, but there are fiber bundles without sections, so $\Gamma(E)=\emptyset$ (for example, take a non-trivial principal bundle, since in the context of principal bundles it is equivalent having a section and being trivial), so the map cannot be surjective. On the other hand, can you see how the map $\Gamma(E)\rightarrow E_x$ need not be injective?