From Husemöller's 'Fiber Bundles' (slightly rephrased):
Proposition: Consider a bundle $\xi: E \to B$, and a mapping f: B' \to B. Then for any $s \in \Gamma(\xi)$ there is a \sigma: B' \to E \times_B B' defined by \sigma(b') = (b', sf(b')) that is a section of $f^*(\xi)$, and $f_\xi \sigma = s f$. If $f$ is an identification map and if $\sigma \in \Gamma(f^*(\xi))$ is such that $f_\xi \sigma$ is constant on $f^{-1}(b)$ for all $b \in B$, then there is a $s \in \Gamma(\xi)$ such that $sf = f_\xi \sigma$.
Proof: [...] For the second statement, we have a factorization of $f_\xi \sigma$ by $f$, giving a map $s: B \to E$ with $sf = f_\xi \sigma$. Moreover, $\xi sf = \xi f_\xi \sigma = f f^*(\xi) \sigma = f$ and $\xi s = 1_B$, since $f$ is surjective. Then $s$ is the desired cross section.
What does 'an identification mapping' mean? Does it merely mean a surjection? I'm sure it can't mean a homeomorphism or even a bijection because that would make the second part trivial.