Let $S$ be a (complex) projective surface and let $C,C'\subset S$ two closed irreducible curves (namely two prime Weil divisors). It is well defined the scheme $C\cap C'$ (as fibered product of $C\rightarrow S$ and $C'\rightarrow S$) which is a closed subscheme of $X$ but I have some problems to understand the sheaf indicated as $\mathcal O_{C\cap C'}$.
1) Some books say that $\mathcal O_{C\cap C'}$ is the "skyscraper sheaf" supported by the set theoretic intersection $|C|\cap |C'|$. But what is this sheaf? Can you define it please? (For example what are its global sections?)
2) Why if $x\in C\cap C'$, then the local ring $\left(\mathcal O_{C\cap C'}\right)_x$ is equal to $\frac{\mathcal O_{X,x}}{(f,g)}$ where $f,g$ are local equations for $C$ and $C'$.
Many thanks in advance.
First, a note about terminology. If $\iota: Y \hookrightarrow X$ is a closed subscheme and $\mathcal{F}$ is a sheaf of $\mathcal{O}_Y$-modules on $Y$, it makes no difference whether we compute $H^i(Y, \mathcal{F})$ or $H^i(X, \iota_* \mathcal{F})$. This "trick" is used so often that it pays to be a little sloppy with the notation and not write $\iota_*$. Sometimes we say "view it as a sheaf on the bigger space" to mean the same thing: take $\iota_*$ of the sheaf.
That is what is happening in your notation: $\mathcal O_{C \cap C'}$ is just the structure sheaf of $C \cap C'$, viewed as a sheaf on $X$. To be precise, one would write it with an $\iota_*$. Since $C \cap C'$ is a bunch of points, this is also sometimes called a skyscraper sheaf. So if $V$ is an open subset of $S$, we have $$\iota_*\mathcal O_{C \cap C'}(V) = \mathcal O_{C \cap C'}(V \cap C \cap C') ``=" \{\text{ functions on } C \cap C' \cap V\}.$$
EDIT: I missed your second question. This is the definition of closed immersion: functions on the closed subscheme all arise as restrictions of functions on the ambient scheme. For any affine $U$ containing $x$, there is a surjective "restriction map" $\mathcal{O}_X(U) \to \mathcal{O}_{C \cap C'}(U)$, and to say that $f$ and $g$ are local equations for $C$, $C'$ is to say that (for a judicious choice of $U$) $f$ and $g$ generate the kernel of this map (more precisely, $f$ generates the kernel of $\mathcal{O}_X(U) \to \mathcal{O}_C(U)$, similarly for $g$, and then use the definition of the fiber product. Note that in this parenthetical remark I am using the "abuse of language" I referred to in the first paragraph of this answer).