Let $X,Y\subset\Bbb{P}^3$ be a smooth surface of degree $d$ and a plane respectively. If $X\cap Y=A+B$ for curves $A,B$, then $(A\cdot B)_X=(A\cdot B)_Y$ (intersection numbers in $X$ and $Y$ respectively).
Considering that $Y\simeq \Bbb{P}^2$, we can see $A,B$ as curves in $\Bbb{P}^2$ having degrees $a,b$ with $a+b=d$, so by Bézout's theorem, $(A\cdot B)_Y=ab$.
For $X$, I'm trying to use the definition $(A\cdot B)_X=\sum_{x\in A\cap B}m_x^X(A\cap B)$, where $m_x^X(A\cap B):=\dim_k \mathcal{O}_{X,x}/(f,g)$, for local parameters $f, g\in\mathcal{O}_{X,x}$ for $A,B$ respectively.
I guess I should be able to show that $m_x^X(A\cap B)=m_x^Y(A\cap B)$, but I don't know how to do that.
There's also the definition $(A\cdot B)_X=h^0(X,\mathcal{O}_{A\cap B})$, where $\mathcal{O}_{A\cap B}:=\mathcal{O}_X/\left(\mathcal{O}_X(-A)+\mathcal{O}_X(-B)\right)$, but I also can't see how's that useful.
Any suggestions?
Let me give two arguments.
Local argument (assuming $A$ and $B$ have no common components)
The curves $A$ and $B$ have natural scheme structures (as complete intersections of type $(1,a)$ and $(1,b)$, respectively). As such they are Cohen--Macaulay schemes. Consequently, they are Cartier divisors on any smooth surface containing them and their intersection numbers $(A\cdot B)_X$ and $(A \cdot B)_Y$ are both equal to the length of the schematic intersection $A \cap B$, which is independent on the embedding into a surface.
Global argument
As $A$ is a complete intersection curve, its normal bundle equals $N_{A/\mathbb{P}^3} = (\mathcal{O}(1) \oplus \mathcal{O}(a))\vert_A$, hence $\deg(N_{A/\mathbb{P}^3}) = (1 + a)a$. From the exact sequence $$ 0 \to N_{A/X} \to N_{A/\mathbb{P}^3} \to N_{X/\mathbb{P}^3}\vert_A \to 0 $$ and the isomorphism $N_{X/\mathbb{P}^3} \cong \mathcal{O}(d)\vert_A$ it follows that $$ A^2_X = \deg(N_{A/X}) = (1+a)a - da = (1 - b)a. $$ Similarly, $B^2_X = (1-a)b$. Therefore $$ 2(A\cdot B)_X = (A+B)^2_X - A^2_X - B^2_X = d - (1-b)a - (1-a)b = 2ab, $$ as required.