It may be a bit unfair to compare $X=mathbb P^1 times mathbb P^1$ to $mathbb P^1$. EDIT: I removed a too optimistic statement about restricting vector bundles on $mathbb P^3$ to a smooth hypersurface, which works for $mathbb P^4$ but not always for $mathbb P^3$.
On the other hand, if we compare $X$ with the smooth surface $mathbb P^2$, then there is this famous result by Horrocks, which says that a vector bundle $E$ on $mathbb P^2$ splits if
$$oplus_{iin mathbb Z} H^1(mathbb P^2,E(i)) =0 $$
Interestingly, over $X$, the same result works. In other words, $E$ on $X$ splits if:
$$oplus_{iin mathbb Z} H^1(X,E(i)) =0 $$
the reason is such $E$ corresponds to a (graded) maximal Cohen-Macaulay (MCM) module over the cone of $X$, namely the hypersurface $R=k[x,y,u,v]/(xu-yv)$. But $R$ has finite MCM type and all the indecomposables have rank one (but note that they will not always be twists of the trivial line bundle). A lot more details and references are available in this paper by Buchweitz, Greuel and Schreyer.
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