I'm asking this question with reference to an answer by BlaCa to an old MathOverflow question I saw here.
We first recall that the $n$-th Hirzebruch surface $\mathbb{F}_n$ can be viewed as the projective bundle $\mathbb{P}(\mathcal{O}_{\mathbb{P}^1} \oplus \mathcal{O}_{\mathbb{P}^1}(n))$. It contains a unique irreducible curve $E$ with self-intersection $-n$. Let $F$ be a fiber of $\mathbb{F}_n \rightarrow \mathbb{P}^1$.
What I'm confused about is the line "The linear system $|E+nF|$ has degree zero on $E$, degree one on $F$ and degree $n$ on $E+F$." What does it mean for a linear system to have some degree over a divisor? Also, how are the degrees zero, one and $n$ deduced?
Moreover, the answer followed up by saying $E$ contracts to a unique singular point whose hyperplane section is a rational normal curve of degree n. Why is this a consequence?
You working over on a surface $\mathbf{F}_n$ so a the divisors $E,F$ are given by (smooth) curves on $\mathbf{F}_n$!
Now what you do is you restrict the divisor $E+ nF$ to $E$ or $F$ and then recall that the associated line bundle to the divisor $E+nF$ becomes after restriction $\mathcal{O}(E+nF)|_E$ or $\mathcal{O}(E+nF)|_F$ and line bundles on curves have a degree (it is just the degree of any non-zero rational section of that line bundle!).
Then use the that $F$ (or $E$ or $E+F$) are effective divisors on $\mathbf{F}_n$ and you have the formula $$\deg(\mathcal{O}(E+nF)|_F)=(E+nF).F$$ which is $(E+nF).F=EF+nF^2=1$ which is what you wanted. Analogously for $E$ and $E+F$!