Let $X$ be a surface inside $\mathbb{P}^3$ of degree $d$, that is given as $f=0$ where $f(x, y, z, w)$ is a homogenous polynomial of degree $d$. Let $\tilde{H}$ be a class of a hyperplane section in $\mathbb{P}^3$ and let $H$ be its restriction to $X$. I want to prove that $H^2=d$. It contradicts my intuition because of the following reason: it seems that $H=\tilde{H} \cap X$ should be a curve of degree $d$ on $X$ and, therefore, by Bezout's theorem the intersection of two general divisors linearly equivalent to $H$ should consist of $d^2$ points meaning that $H^2=d^2$ instead of $H^2=d$.
What is wrong with my intuition?
Let us work "by hand" the first nontrivial example, namely $d=2$. We fix $$ X= V(x_0x_3-x_1x_2) \subset \mathbb{P}^3. $$ As it is well known, we can identify $X$ with $\mathbb{P}^1 \times \mathbb{P}^1$, using the isomorphism $$ \Phi\colon \mathbb{P}^1 \times \mathbb{P}^1 \to X, \qquad [y_0, y_1], [z_0, z_1] \mapsto [y_0z_0, y_0z_1, y_1z_0, y_1z_1]. $$ Now, consider some hyperplane $\overline{H}=V(F)$, where $F=a_0x_0+a_1x_1+a_2x_2+a_3x_3$. The intersection $\overline{H} \cap X$ is then computed as $$ \overline{H} \cap X = V(\Phi^\ast F)=V(a_0y_0z_0+a_1y_0z_1+a_2y_1z_0+a_3y_1z_1). $$ This is, $H \subset \mathbb{P}^1 \times \mathbb{P}^1$ (identifying $H$ with its preimage under $\Phi$) is a curve of bidegree $(1,1)$ inside $\mathbb{P}^1 \times \mathbb{P}^1$.
Now, it is not so difficult to convnince ourselves that if $H, H'$ are "sufficiently nice" $(1,1)$ curves, then $H \cdot H'=2$. For instance, if $H=V(x_0y_0)$ and $H'=V(x_1y_1)$, then $H \cdot H'=2$ by counting the intersections.