If we have that $\alpha,\beta$ are isomorphic bundles, say $p_1:E_1\to B_1$ and $p_2:E_2\to B_2$ respectively, then there is a map $f:B_1\to B_2$ that is the identity map, and we get that:
$$\begin{array}{ccc}E_1& \overset{F}{\to} &E_2\\ \downarrow&&\downarrow\\B_1&\overset{f}{\to}&B_2\end{array}$$
Is the pullback bundle $f^*E_2$ the same thing as $E_1$?
It seems we obtain $f^*E_2$ (which is completely a notational choice right) where $$f^*E_2=\{(b,e):b\in B_1,p_2(e)=f(b)\}$$ Now since $f$ is the identity, this pullback bundle has in it's total space elements $(b,e)$ where $p_2(e)=b$, and we can see that the square of this pullback bundle commutes, since $$\pi_2(F((b,e)))=\pi_2(e)=b=f(b)=f(\pi_1((b,e))$$ and so this pullback bundle is isomorphic to $\beta$ and thus isomorphic to $\alpha$ if the linear isomorphisms of fibers also holds, is that logic correct so far?