I'm trying to understand Fulton's approach to the question how many plane conics in $\mathbb P^3$ meet 8 general lines. Suppose $V$ is a 4-dimensional vector space, and $\mathbb P^3 = \mathbb P(V)$. Then the Grassmannian $G = G(V, 3)$ parametrizes planes in $\mathbb P(V)$, and comes with a canonical sequence of locally free sheaves $$ 0 \to \mathbb A \to V \otimes \mathcal O_G \to \mathbb B \to 0,$$ with $\operatorname{rk}(\mathbb A) = 1$ and $\operatorname{rk}(\mathbb B) = 3$. Plane conics are parametrized by the $\mathbb P^5$-bundle $\mathbb P(\operatorname{Sym}^2\mathbb B^\vee) \to G$. Suppose $l \subset \mathbb P(V)$ is a line which corresponds to a 2-dimensional quotient $V \to W$. Then the planes which contain $l$ form a dual line $l^\vee \subset G$, and on $U = G \setminus l^\vee$ one has the two short exact sequences \begin{equation}\require{AMScd}\begin{CD} 0 @>>>\mathbb A|_U @>>> V \otimes \mathcal O_U @>>> \mathbb B|_U @>>> 0 \\ {}@VVV @VVV @VVV\\ 0 @>>>\mathbb A|_U @>>> W \otimes \mathcal O_U @>>> \mathcal L @>>> 0 \\ \end{CD}\end{equation} for some line bundle $\mathcal L$ on $U$. Taking second symmetric power of the surjection $\mathbb B|_U \to \mathcal L$ and combining this with the canonical quotient $\pi^* \operatorname{Sym}^2 \mathbb B^* \to \mathcal O_{\pi}(1)$ one obtains a global section $$ s \in \Gamma(\pi^{-1} U, \pi^* \operatorname{Sym}^2 \mathcal L \otimes \mathcal O_{\pi}(1)).$$ By construction this section vanishes exactly at those conics which meet $l$.
Fulton claims that $\mathcal L = \mathbb A|_U$. Here is what I tried:
We have $G \cong \mathbb P^3$, and with this identification $\mathbb A = \mathcal O(1)$, and so $\operatorname{Pic}(G) = \mathbb Z \cdot \mathbb A$. Also $\operatorname{Pic}(G) \cong A_3(G)$, since $G$ is smooth, and looking at the exact sequence $$ 0 = A_3(l^\vee) \to A_3(G) \to A_3(U) \to 0$$ we see that $\mathcal L = \mathcal O(k)|_U$ for some uniquely determined $k$.
But why is $k=1$? I think if $k > 0$ then any $\mathcal O(k)$ is generated by $x_0^k$ and $x_1^k$ outside the line $\{x_0 = x_1 = 0\}$, so the fact that $\mathcal L$ is generated by two elements is not sufficient.