As explained in abx's comment, the canonical divisor of your surface is given by $$K_X=-3 \pi^*L + \sum E_i,$$ and this is precisely the class of $-\widetilde{C}_1$ (note that $C_1$ is a class in $\mathbb{P}^2$, I put the tilde to specify that we are considering the strict transform in $X$).  

Now, $\widetilde{C}_1$ is the strict transform of a smooth cubic curve: since the self-intersection number of a cubic is $9$ and you are blowing up nine point on it, the self intersection of the strict transform is $(C_1)^2=9-9=0,$ as claimed.

Finally, the strict transform of your pencil of cubics is a base-point free pencil of elliptic curves in $X$, providing an elliptic fibration $\pi \colon X \to \mathbb{P}^1$. By construction, the curve $\widetilde{C}_1$ is a fibre of $\pi$, hence it is a "vertical divisor", after all.