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 $(n-2)$th Hirzebruch surface.



A specific degeneration of a rank 2 bundle
Let $F = O \oplus O(n-2,0)$ and denote the line $P^1\times[0:1]$ by $L$. Note that $F\_{|L} = O \oplus O(n-2)$. Consider any surjective map $O\_L \oplus O\_L(n-2) \to O\_L(n-1)$ (for example the one given by $u^{n-1}$ on the first summand and by $v$ on the second, where $(u:v)$ are the homogeneous coordinates on $L$). Consider the composition $F \to F\_{|L} \to O\_L(n-1)$ and let $E$ be its kernel. Note that for any line $L' = P^1\times[x:y]$ we have $E\_{|L'} = F\_{|L'} = O \oplus O(n-2)$. On the other hand, restricting to $L$ we obtain an exact quadruple
$$
0 \to L\_1i^\*O\_L(n-1) \to E\_{|L} \to O\_L \oplus O\_L(n-2) \to O\_L(n-1) \to 0
$$
where $i:L \to P^1\times P^1$ is the embedding.
Note that $L\_1i^\*O\_L(n-1) = O\_L(n-1)$, and the kernel of the rightmost map is $O\_L(-1)$. Hence we have an exact triple
$$
0 \to O\_L(n-1) \to E\_{|L} \to O\_L(-1) \to 0.
$$
Since there are no notrivial extensions, we see that $E\_{|L} = O\_L(-1) \oplus O\_L(n-1)$.
So, the bundle $E$ gives what you need.


