I wish to know if there is a rank 2 vector bundle $E$ on $\mathbb{P}^1 \times \mathbb{P}^1$ such that $\mathbb{P}(E)$ when restricted to $\mathbb{P}^1 \times [0:1]$ is the $n$th Hirzebruch surface and when restricted to $\mathbb{P}^1 \times [x:y]$ is the $(n2)$th Hirzebruch surface.
Let $F = O \oplus O(n2,0)$ and denote the line $P^1\times[0:1]$ by $L$. Note that $F_{L} = O \oplus O(n2)$. Consider any surjective map $O_L \oplus O_L(n2) \to O_L(n1)$ (for example the one given by $u^{n1}$ on the first summand and by $v$ on the second, where $(u:v)$ are the homogeneous coordinates on $L$). Consider the composition $F \to F_{L} \to O_L(n1)$ and let $E$ be its kernel. Note that for any line $L' = P^1\times[x:y]$ we have $E_{L'} = F_{L'} = O \oplus O(n2)$. On the other hand, restricting to $L$ we obtain an exact quadruple $$ 0 \to L_1i^*O_L(n1) \to E_{L} \to O_L \oplus O_L(n2) \to O_L(n1) \to 0 $$ where $i:L \to P^1\times P^1$ is the embedding. Note that $L_1i^*O_L(n1) = O_L(n1)$, and the kernel of the rightmost map is $O_L(1)$. Hence we have an exact triple $$ 0 \to O_L(n1) \to E_{L} \to O_L(1) \to 0. $$ Since there are no notrivial extensions, we see that $E_{L} = O_L(1) \oplus O_L(n1)$. So, the bundle $E$ gives what you need. 

