Let $S\subset\Bbb{P}^3_\Bbb{C}$ be a nonsingular cubic algebraic surface containing a line $L$. We define the projection $\varphi:S\setminus L\to \Bbb{P}^1$ away from $L$ in the following manner. Take $M=\Bbb{P}^1\subset\Bbb{P}^3$ a line disjoint from $L$. For $P\in S\setminus L$, define $\varphi(P)$ as the only point in the intersection of $M$ and the plane $\Pi$ containing $P$ and $L$.
In Miles Reid's chapters on algebraic surfaces (page $9$), there's the following reasoning for extending the projection $\varphi$ to the whole surface $S$ (I paraphrase):
Let $H=L+F$ be a hyperplane section containing $L$. Computing the intersection numbers we get $HL=1$, $FL=FH=2$, so $H^2=3$ and $F^2=0$. Since the $F$ are effective and disctinct, they must be disjoint, so $\varphi$ is well-defined along $L$ as well".
I do understand the computations of the intersection numbers and also the fact that $F\geq 0$ and $F^2=0$ implies that the linear system $|F|$ is base-point free, which means there is a morphism $S\to\Bbb{P}^n$, where $n=\dim|F|$ (dimension as a projective space).
I still wasn't able to explain 1) why $n=1$ and 2) why the new morphism $S\to\Bbb{P}^1$ given by $|F|$ actually extends $\varphi$.
Consider the short exact sequence $$ 0 \to \mathcal{O}_S(-L) \to \mathcal{O}_S \to \mathcal{O}_L \to 0. $$ Twisting it by $H$ one gets $$ 0 \to \mathcal{O}_S(F) \to \mathcal{O}_S(H) \to \mathcal{O}_L(1) \to 0. $$ The global sections are 4-dimensional for the middle term, 2-dimensional for the right term, and the restriction morphism is surjective, hence the global sections of $\mathcal{O}_S(F)$ are 2-dimensional.
Moreover, this proves that the linear subsystem in $|H|$ consisting of divisors containing $L$ (which defines the morphism $\varphi$ away from $L$) coincides with the linear system $|F| + L$. If you recall the definition of a morphism given by a linear system you will see that the morphisms of $S \setminus L$ given by these two linear systems coincide.