Let $V_1, V_2, W_1, W_2$ be finite dimensional vector spaces over an algebraically closed $K$ field and $r_1, r_2$ positive integers. For $i \in \{1,2\}$, let $X_i$ be the set of $K$-linear maps $W_i \to V_i$ of fixed rank $r_i$. Let $Y$ be the set of quadruples $(f_1, f_2, g, h)$ with $(f_1, f_2) \in X_1 \times X_2$ and $(g, h) \in \operatorname{Hom}_K(V_2, V_1) \times \operatorname{Hom}_K(W_2, W_1)$ such that $gf_2 = f_1h$. Let $\pi \colon Y \to X_1 \times X_2$ be the projection. We assume all spaces to carry the Zariski topology.
Then $X_1$ and $X_2$ are irreducible and for all $(f_1, f_2) \in X_1 \times X_2$, the fiber $\pi^{-1}(f_1, f_2)$ is a finite dimensional vector space of constant dimension.
I struggle to show that $\pi$ is locally trivial. I tried to choose direct complements of $\operatorname{Im}f_i$ and $\operatorname{Ker}f_i$ to get a suitable open neighborhood of $(f_1, f_2) \in X_1\times X_2$, but that didn't help...
The group $G = GL(V_1) \times GL(V_2) \times GL(W_1) \times GL(W_2)$ acts naturally on $Y$ and $X = X_1 \times X_2$, the map $\pi$ is equivariant, and the action on $X$ is transitive. This implies that $\pi$ is locally trivial.
Indeed, choose a point $x_0 \in X$, let $H \subset G$ be the stabilizer of $x_0$ and let $Y_0 = \pi^{-1}(x_0)$. Then $Y_0$ is a representation of $H$. Moreover, there is a map $$ \phi \colon G \times Y_0 \to Y, \qquad (g,y) \mapsto g \cdot y. $$ The group $H$ acts on $G \times Y_0$ via $$ (g,y) \mapsto (g \cdot h^{-1}, h \cdot y), $$ and the map $\phi$ is invariant for this action. Therefore it induces a map $$ G \times^H Y_0 := (G \times Y_0)/H \to Y, $$ and it is easy to see it is an isomorphism.
Finally, it is easy to see that the projection $$ (G \times Y_0)/H \to G/H $$ is a locally trivial vector bundle with fiber $Y_0$.