On page 101 of Lang's "Differential and Riemannian Manifolds", he introduces fibre-bundles because composing vector bundles does not give vector bundles.
To be more precise, given a vector bundle $p : E \rightarrow X$ and the tangent bundle $p: TE \rightarrow E$, he states that the composite $f = p \circ \pi$ is "only a fibre bundle over $X$, a fact which is obvious by picking trivializations in charts".
Looking at the relevant trivializations, this is not clear to me at all. Why is $f$ not a vector bundle in general?
If $p\colon E \to X$ and $q\colon E' \to E$ are vector bundles, then the composite $qp\colon E' \to X$ is also a vector bundle. To see this, let $s_0\colon X \to E$ denote the zero-section of $p$, and let $q_X\colon E'_X \to X$ denote the pullback of $q$ along $s_0$, which is a vector bundle over $X$. Pulling $E'_X$ back along $p$, we obtain a vector bundle $E'_X \times_X E \to E$ over $E$. I claim that there is an isomorphism of vector bundles \begin{aligned} E' \cong E'_X \times_X E. \end{aligned} Indeed, recall that if $H\colon Y \times [0,1] \to Z$ is a homotopy, then there is for every vector bundle $F$ over $Y$ an isomorphism $H_0^*F \cong H_1^*F$. Applying this to the homotopy $H\colon E \times [0,1] \to E, (e,\lambda) \mapsto \lambda x$, we see that $H_0 = s_0 \circ p\colon E \to E$ while $H_1 = \operatorname{id}_E$. In particular we get isomorphisms \begin{aligned} E' = H_1^*E' \cong H_0^*E' \cong (p \circ s_0)^*E' \cong p^*s_0^*E' = p^*E'_X = E'_X \times_X E. \end{aligned} It thus remains to show that the composite \begin{aligned} E'_X \times_X E \xrightarrow{\operatorname{pr}_2} E \xrightarrow{p} X \end{aligned} is a vector bundle. But this is the well-known direct sum $E'_X \oplus E$ of the vector bundles $E'_X$ and $E$ over $X$.