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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.

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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.

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