The degree of a variety $X$ of dimension $r$ is defined by $r!$ times the leading coefficient of its Hilbert polynomial. This is the defination given in Hartshorne, but I find it is very hard to handle with. If $X$ is a curve, one can cut it by generic hyperplane, and count the number of intersection points. This gives the degree of curve. However, for the surface, I want to understand the following claim used by Hartshorne:
(1) $X$ is a surface with embedding $X \to \mathbb{P}^n$, and $D=O_X(1)$, then the degree of $X$ in $\mathbb{P}^n$ is $D^2$.
(2) $X$ is a surface with embedding $X \to \mathbb{P}^n$, and $D=O_X(1)$. Suppose $h$ is a divisor in $X$, then the degree of $h$ in $\mathbb{P}^n$(viewed as curves) is $D.h$
I certainly wish to see the rigorous proves, but presumably they will be long and not very enlightening. So any intuitive argument is also greatly welcome!
(2) Let $H$ be a generic hyperplane in $P=\mathbb P^n$, then, as invertible sheaves, we have $O_X(1)\simeq O_P(H)|_X$. So $$O_X(1).h:=\deg O_X(1)|_h=\deg O_P(H)|_h$$ is the intersection number of $h$ with $H$.
(1) The degree of $X$ in $P$ is the intersection number of $X$ with a generic linear space of codimension $2$. Or, the intersection number of $X\cap H$ with $H'$, where $H$ and $H'$ are generic. So it is the degree of $X\cap H$ (scheme theoretic intersection) in $P$.
As $O_X(1)$ is generated by its global sections, $O_X(d)\simeq O_X(h)$ for some curve $h$ in $X$ (in fact $h=X\cap H$ is OK for any $H$ which does not contain $X$). So $$D^2=O_X(1).h=\deg O_X(1)|_h=\deg h=\deg X.$$