Given a smooth vector bundle $\pi:E\to M$, the vertical bundle $T_v(E)$ is by definition $\ker T\pi$, which is a subbundle of $T(E)$. It is asserted in this answer that there is a canonical isomorphism $$T_v(E)\cong \pi^* E.$$
Georges Elencwajg writes the following:
Recalling that for a finite-dimensional vector space $V$ (seen as a manifold) we have for each $a\in V$ a canonical isomorphism $T_a(V)=V$ we obtain the canonical isomorphism $T_v(E)=\pi^*(E)$
I am still unable to see how one obtains the claimed isomorphism and would be thankful for elucidation.
The vertical bundle consists of the fibers of $E$ as subspaces of $TE$. (This is because we identify the tangent vectors to a vector space, as Georges said, with the vector space itself.) That is, at a point $\xi\in E$, $(T_vE)_\xi = E_{\pi(\xi)}$. Thus, by definition of the pullback bundle, $T_vE = \pi^*E$.