Here is our setting: $X$ is a surface (in the sense of chapter V in Hartshorne), $D$ is a reduced curve on $X$ (possibly singular), $C_i$ is an irreducible component of $D$, $f: \tilde{C}_i \rightarrow C_i$ is the normalization map. Hartshorne writes
$\deg f^* (\mathscr{L}(D) \otimes \mathcal{O}_{C_i})$ is just $D. C_i$, because we can represent $\mathscr{L}(D)$ as a difference of nonsingular curves meeting $C_i$ transversally, so this degree is preserved by $f^*$.
I can see that $D \sim E-E'$, where $E$ and $E'$ are nonsingular curves on $X$ meeting $C_i$ transversally. However, how does this give that $\deg f^* (\mathscr{L}(D) \otimes \mathcal{O}_{C_i}) = D. C_i$? Many thanks.
Best Answer
Since pullback commutes with tensor products and duals, and intersection product is linear and respects rational equivalence, it suffices to prove that $\deg f^*(\mathcal{L}(E)\otimes\mathcal{O}_{C_i})=E.C_i$ for $E$ a nonsingular curve meeting $C_i$ transversely.
On the one hand, $E.C_i$ is just $\#(E\cap C_i)$ since they meet transversely (cf your previous question).
On the other hand, since $E\cap C_i$ is a divisor supported on the regular locus of $C_i$, the pullback of this via $f$, which is an isomorphism on the regular locus, doesn't change. So computing the degree is still the same thing as computing the number of points in the divisor, which didn't change!