For a fiber bundle $\pi: E\rightarrow B$ with typical fiber a manifold $F$, we define the vertical subbundle $V$ by the exact sequence of verctor bundles over $E$: $$0\rightarrow V\rightarrow TE \rightarrow \pi^*TB\rightarrow0.$$ In the case when $F$ has a Lie group structure and $E$ is a principal bundle, we know that $V$ is the trivial bundle with fiber $T_eF$, the Lie algebra of $F$.
In general, can we say anything about $V$ and $TF$?
It's not trivial in general. For example, the Klein bottle $K$ is the total space of a fibre bundle $S^1 \to K \xrightarrow{\pi} S^1$. Note that $\pi^*TS^1$ is trivial and if $V$ were isomorphic to $TS^1$, it would also be trivial. This would imply that $TK$ is trivial, but this is not the case as $K$ is non-orientable.
Let $i : F \to E$ be the inclusion of a fiber. Since $\pi\circ i$ is constant, we see that $0 = (\pi\circ i)_* = \pi_*\circ i_*$, so $i_*TF \subseteq \ker \pi_* = V$. By comparing ranks, we see that in fact $i_*TF = V$ and hence $TF \cong i^*V$.