Intersection numbers in $\mathbb{P}^1$-bundles

Here are the details.

Let $L_p\cong\mathbb{P}^1$ be a line in $\mathbb{P}^2$ through $p$. Your exact sequence restricted to $L_p$ yields the following exact sequence $$0\rightarrow\mathcal{O}_{\mathbb{P}^1}(1)\rightarrow\mathcal{E}_{|L_p}\rightarrow\mathcal{O}_{\mathbb{P}^1}(-2)\rightarrow 0 $$ and hence $\mathcal{E}_{|L_p}\cong\mathcal{O}_{\mathbb{P}^1}(1)\oplus\mathcal{O}_{\mathbb{P}^1}(-2)$. Therefore, $\pi^{-1}(L_p) = \mathbb{P}(\mathcal{O}_{\mathbb{P}^1}(1)\oplus\mathcal{O}_{\mathbb{P}^1}(-2))$ is the Hirzebruch surface $\mathbb{F}_3$. Let $\overline{\xi}$ be the class of the only curve with self-intersection $-3$ on $\mathbb{F}_3$, and let $f$ be the class of a fiber of $\pi_{|\mathbb{F}_3}:\mathbb{F}_3\rightarrow\mathbb{P}^1$. Then if $\widetilde{H} = \pi^{*}\mathcal{O}_{\mathbb{P}^2}(1)$ and $\xi = c_1(\mathcal{O}_X(1))$ we have $\widetilde{H}_{|\mathbb{F}_3} = f$ and $\xi_{|\mathbb{F}_3} = f+\overline{\xi}$.

Since $c_1(\mathcal{E})=-1$ and $c_2(\mathcal{E}) = 1$ the canonical divisor of $X$ is given by $-K_X = 4\widetilde{H}+2\xi$ and we may compute the intersection $-K_W\cdot\overline{\xi}$ inside $\mathbb{F}_3$, that is $$-K_X\cdot\overline{\xi} = (4\widetilde{H}+2\xi)_{|\mathbb{F}_3}\cdot\overline{\xi} = (4f+2(f+\overline{\xi}))\cdot\overline{\xi} = (6f+2\xi) = 6+2(-3) = 0.$$

It seems $C$ is not well-defined. Let $\ell$ be the line such that $\pi(C) = \ell$. Then we have the $\mathbb{P}^1$-bundle $$ \pi'\colon\mathbb{P}(\mathcal{E}|_\ell) \rightarrow \ell. $$ Any section of this bundle will give rise to a curve $C$ as you defined (and indeed, such $\mathbb{P}^1$ bundles admit infinitely many sections). The intersection number $-K_X \cdot C$ will depend on which section you choose.