Let $S$ be an algebraic complex surface, and $D=[(U_\alpha,f_{\alpha})]$ is a Cartier divisor over $S$, and let $\cal{O}_S(D)$ be the sheaf associated to $D$. And let $C$ be a complex compact curve in $S$, and let $\iota : C \to S$ be the inclusion map. Let's denote by $\cal{O}^D=\cal{O}_S/\cal{O}_S(-D)$. I use this notation to avoid the confusion with $\cal{O}_C$ the sheave of holomorphic functions on $C$. We have the two exact sequences :
$$0 \to \cal{O}_S(-C) \to \cal{O}_S \to \cal{O}^C \to 0,$$
$$0 \to \cal{O}_S(-C-D) \to \cal{O}_S(-D) \to \cal{O}^C \otimes \cal{O}_S(-D) \to 0.$$
Then $$(C.D)=\chi(\cal{O}_S)-\chi(\cal{O}_S(-C))-\chi(\cal{O}_S(-D))+\chi(\cal{O}_S(-C-D))= \chi(\cal{O}^C)-\chi(\cal{O}^C \otimes \cal{O}_S(-D)).$$
In the book of Chris Peters "AN INTRODUCTION TO COMPLEX ALGEBRAIC GEOMETRY : WITH EMPHASIS ON THE THEORY OF SURFACES" he claim that by using Riemann-Roch theorem we get: $$(C.D)=deg(D_{/C}),$$
where $D_{/C}(x)=ord_x((f_{\alpha})_{/C}))$, and $\alpha$ such that $x \in U_{\alpha}$.
First of all, I don't understand this claim.
Maybe we should prouve that there existe an isomorphism between $\iota_{*}(\cal{O}_C)$ and $\cal{O}^C$, and an isomorphism between $\cal{O}^C \otimes \cal{O}_S(-D)$ and $ \iota_{*}(\cal{O}_C(-D_{/C}))$ to answer this question, but I don't have an idea to prove it. In this case, we can conclude by using the fact $$\mathrm{H}^i(C,\cal{F})\simeq \mathrm{H}^i( S,\iota_{*}\cal{F}).$$
I'll appreciate any help.
First remark: what you're referring to $\mathcal{O}^C$ is in fact $\mathcal{O}_C$ (why?).
Now, coming to the question: you already have shown that $(C.D) = \chi(\mathcal{O}_C) - \chi(\mathcal{O}_C(-D))$. Now, Riemann-Roch for curves says that for $L$ a line bundle on $C$, $\chi(L) = \deg L - g+1$. So, $$\begin{align}(C.D)&= \chi(\mathcal{O}_C) - \chi(\mathcal{O}_C(-D))\\ &= (0-g+1)-(\deg \mathcal{O}_C(-D) - g+1) \\ &= -\deg \mathcal{O}_C(-D) = \deg \mathcal{O}_C(D).\end{align}$$