Suppose that $X$ is a scheme with a $B$-action, and that $\lambda: B \to \mathbb{C}^*$ is a character of $B$. We then have a geometric vector bundle $\pi: X \times^B k \to X/B$, where $B$ acts on $X \times^B k$ via $(x,v).b = (xb,\lambda(b)^{-1}v)$. Suppose that $X'$ is a scheme with a $B'$-action and that $f: X'/B' \to X/B$ is a morphism of schemes. Then the pullback bundle is $(X'/B') \times_{X/B} (X \times^Bk)$ with projection onto the first factor.
If this was merely a topological vector bundle, then we could easily identify the total space of the pullback bundle - it is just $\{(x',(x,v)) \mid f(x') = \pi(x,v) \}$. But since this is a geometric bundle, we are in the category of schemes, and the fibre product in Schemes is not the same as in the category of topological spaces or sets.
Is it nonetheless true that as a set the total space $(X'/B') \times_{X/B} (X \times^Bk)$ admits the 'easy' description available in the case of topological bundles?