Let $\pi :E\to M$ be a vector bundle with typical fiber $V$. Suppose that $\pi^*E$ is the pullback bundle of $E$ by $\pi$. If $(\zeta, \xi) \in \pi^*E$, then the map $\pi(\zeta +t\xi)$ is constant in $t$, because both $\zeta$ and $\xi$ are in the same fiber, call it $E_p$. In my textbook (Differential Geometric Structures by Poor), it is said that the map $\mathcal J:\pi^*E\to TE$ given by $$\mathcal J(\zeta +t\xi) = \frac d{dt}\Big|_{t=0}(\zeta +t\xi)$$ actually maps into the vertical bundle $\mathcal VE$. My questions are:
- How is $\frac d{dt}\Big|_{t =0} (\zeta + t\xi) \in TE$? And
- How is it in $\mathcal VE$?
As you noted, the elements of $\pi^*E$ are exactly the pairs $(\zeta,\xi)\in E_p\times E_p$ for some point $p\in M$.
Thus, for any $(\zeta,\xi)\in \pi^*E$ the image of the curve $\gamma:\Bbb R\to E,\ \ t\mapsto \zeta+t\xi\ $ lies fully in the fiber $E_p$ (where $p=\pi(\zeta)=\pi(\xi)$).
So, $\pi(\gamma(t))=p$ is constant, therefore $$d(\pi\circ\gamma)\,(0)\ =\ d\pi\,d\gamma\,(0) = 0\,,$$ i.e. the differential (speed vector) of $\gamma$ at $t=0$ (or at any $t$ actually) will be in the vertical bundle $\mathcal V_pE\,\subseteq T_pE\,\subseteq TE\ $ (and can be naturally identified with $\xi\in E_p$, by the way).