Why do we retain exactness when tensoring by $\mathcal{O}_C$ in Hartshorne, Lemma V.1.3?

Just to answer your question as posed, the sequence remains exact because $C$ and $D$ meet transversally, in particular, $C \cap D \subsetneq C$. In particular, the local equation for $D$ remains nonzero when restricted to $C$. Since $C$ is integral, the map $\mathcal{O}_X(-D) \to \mathcal{O}_X$ (multiplication by the local equation for $D$) remains injective after tensoring with $\mathcal{O}_C$.

If you want to see this algebraically: the sequence

$$0 \to \mathcal{O}_X(-D) \to \mathcal{O}_X \to \mathcal{O}_D \to 0$$

corresponds (in a sufficiently small affine chart of $X$, say $X = \mathrm{Spec}(R)$, where $D$ has local equation $f$ and $C$ has local equation $g$,) to the sequence

$$0 \to R \to R \to R/(f) \to 0,$$

where the first map is multiplication by $f$. Tensoring with $C$ means tensoring with $R/g$, as you said:

$$0 \to R/(g) \to R/(g) \to R/(f,g) \to 0,$$

where the first map is again multiplication by $f$ and not, a priori, injective anymore. But by our hypothesis, $C \cap D \subsetneq C$, so the image of $f$ is not zero in $R/g$. Since $R/g$ is a domain, this means multiplication by $f$ is still injective.

Note that $$0 \to \mathcal{O}_X(-D) \to \mathcal{O}_X \to \mathcal{O}_D \to 0$$ tensor by $\mathcal{O}_C$ is just

$$0\to {\mathcal Tor}_1( \mathcal{O}_D,\mathcal{O}_C)\to \mathscr{L}(-D) \otimes \mathcal{O}_C \to \mathcal{O}_C \to \mathcal{O}_{C \cap D} \to 0$$ since $\mathcal{O}_X$ is flat.

Now ${\mathcal Tor}_1( \mathcal{O}_D,\mathcal{O}_C)$ is a sheaf supported on ${C \cap D}$ and $\mathscr{L}(-D) \otimes \mathcal{O}_C$ is locally free on $C$.

Note that the subsheaf of a locally free sheaf can not be torsion, hence $${\mathcal Tor}_1( \mathcal{O}_D,\mathcal{O}_C)=0$$.

I actually was confused by large portions of V.I, in particular, this part. Hopefully someone will come by and tell us why we're being silly.

I'd like to give you an alternate path to proving the various basic theorems here, that I personally found more helpful. I would have made this a comment, but, unfortunately, it's too long.

Hopefully it's of some help for you, it's the only way I remember how the stuff in V.I works!

In the usual way, define the multiplicity of the intersection of two non-singular curves $C$ and $D$ on a surface $X$ at a point $x$ to be $m_x(C,D):=\ell_{\mathcal{O}_{X,x}}(\mathcal{O}_{X,x}/(f,g))$ where $f$ and $g$ are local equations defining $C$ and $D$ at $x$. Define then $\langle C,D\rangle$ to be $\displaystyle \sum_{x\in C\cap D}m_x(C,D)$. Clearly then $m_x(C,D)=\#(C\cap D)$ if $C$ and $D$ meet transversely. Moreover, it's clear that $m_x(C,D)=\dim_k H^0(X,\mathcal{O}_{C\cap D})$.

Now, for any two line bundles $\mathscr{L},\mathscr{L}'$ define $\langle\mathscr{L},\mathscr{L}'\rangle$ by

$$\chi_X(\mathcal{O}_X)-\chi_X(\mathscr{L}^{-1})-\chi_X(\mathscr{L}'^{-1})+\chi_X(\mathscr{L}^{-1}\otimes\mathscr{L}'^{-1})$$ I claim then that:

For $C$ and $D$ as above, $\langle C,D\rangle=\langle\mathcal{O}(C),\mathcal{O}(D)\rangle$.

To prove this, merely check that the following sequence is exact

$$0\to\mathcal{O}(-D-C)\to\mathcal{O}(-C)\oplus\mathcal{O}(-D)\to \mathcal{O}_X\to i_\ast\mathcal{O}_{C\cap D}\to 0$$

where $i:C\cap D\to X$ is the inclusion, and the first two maps are $x\mapsto s'x-sx$, and $(x,y)\mapsto sx+s'y$, where $s$ and $s'$ global sections of $\mathcal{O}(C)$ and $\mathcal{O}(D)$ respectively. This is easy to check affine locally.

Then, taking $\chi$ we get

$$\chi_X(i_\ast(\mathcal{O}_{C\cap D}))=\chi_X(\mathcal{O}_S)-\chi_X(\mathcal{O}(-D))-\chi_X(\mathcal{O}(-C))+\chi_X(\mathcal{O}(-C-D))$$ which gives the desired result after noting that

$$\chi_X(i_\ast\mathcal{O}_{C\cap D})=\chi_{C\cap D}(\mathcal{O}_{C\cap D})=H^0(X,\mathcal{O}_{C\cap D})$$ since $C\cap D$ is zero dimensional $\blacksquare$

Now, we want to prove that for a smooth curve $C\subseteq X$, and a line bundle $\mathscr{L}$ that $$\langle\mathcal{O}(C),\mathscr{L}\rangle=\deg(i^\ast\mathscr{L})$$ where $i$ is the embedding $C\to X$. Indeed, if this were true then, for two smooth transversely interesting curves $C$ and $D$ we have that

$$\#(C\cap D)=\langle C,D\rangle=\langle \mathcal{O}(C),\mathcal{O}(D)\rangle=\deg(i^\ast\mathcal{O}(D))$$ which is what you were after.

To prove this one just considers the closed subscheme SES $$0\to\mathcal{O}(-C)\to\mathcal{O}_X\to i_\ast\mathcal{O}_C\to 0$$ Tensor this with $\mathscr{L}$ to get $$0\to\mathcal{O}(-C)\otimes\mathscr{L}\to\mathscr{L}\to i_\ast\mathcal{O}_C\otimes\mathscr{L}\to 0$$ Taking $\chi_X$ gives $$\chi_X(\mathscr{L}^{-1})-\chi_X(\mathscr{L}^{-1}\otimes \mathcal{O}(-C))=\chi_C(i^\ast\mathscr{L}^{-1})$$ where I used the projection formula to see that $$\mathscr{L}^{-1}\otimes i_\ast\mathcal{O}_C=i_\ast(i^\ast\mathscr{L})$$ and the fact that $\chi_X(i_\ast(i^\ast\mathscr{L}^{-1}))=\chi_C(i^\ast\mathscr{L}^{-1})$.

So,

$$\begin{aligned}\langle\mathcal{O}(C),\mathscr{L}\rangle &=\chi_X(\mathcal{O}_X)-\chi_X(\mathscr{L}^{-1})-\chi_X(\mathcal{O}(-C))-\chi_X(\mathcal{O}(-C)\otimes\mathscr{L}^{-1})\\ &=\chi_C(\mathcal{O}_C)-\chi_C(i^\ast\mathscr{L}^{-1})\\ &=-\deg(\mathscr{L}^{-1})\\ &=\deg(\mathscr{L})\end{aligned}$$ where I have used "Riemman-Roch" (""=no need for Serre duality) at the obvious place.