I'm trying to prove the following (simplified) version of the projection formula
Let $S',S$ be non-singular algebraic surfaces and $\pi:S'\to S$ a birrational map between them. If $D\in \text{Div}(S)$ and $D'\in \text{Div}(S')$, then: $$(\pi^*D)\cdot D'=D\cdot(\pi_*D')$$
By linearity, we may just consider $D, D'$ prime divisors.
Let $X\subset S'$ and $Y\subset S$ closed sets such that $\pi:S'\setminus X\to S\setminus Y$ is an isomorphism.
The claim is obviously true if $D\subset S\setminus Y$ and $D'\subset S'\setminus X$.
Now if $D'\cap X\neq \emptyset$ or $D\cap Y\neq\emptyset$, I don't know what to do.
In the particular case $\pi:S'\to S$ is the blow-up at the point $P\in S$ with $E:=\pi^{-1}(P)$, I was able to deal with some specific divsors. For example, when $D'=E$ and $D=C$ where $C$ is a curve through $P$. If $m$ is the multiplicity of $P$ in $C$, we can use the formula $\pi^*(C)=\widetilde{C}+mE$.
The result for this specific case was not so evident (in my opinion) and I still can't see how the general case should go.
Nine months later, I'll post the answer I was pregnant with.
Since a birrational morphism between smooth surfaces is a composition of blowups and isomorphisms, it is enough to consider $\pi$ as a single blowup.
By linearity, it is enough to consider $D,D'$ prime divisors. We have two cases:
(1) $D'=E$. Let $\widetilde{C}$ be the stric transform of $C$. We have the formula $\pi^*C=\widetilde{C}+mE$, where $m$ is the multiplicity of $P$ in $C$. Therefore $(\pi^*C)\cdot E=\widetilde{C}\cdot E-m=0$. But since $\pi_*E=0$, we get $C\cdot (\pi_*E)=0$ and we're done.
(2) $D'\neq E$. In this case $\pi_*D'=\pi(D')$ and $D'$ is the strict transform of $\pi(D')$, therefore $\pi^*(\pi_*D)=D'+mE$. By the previous case, $(\pi^*C)\cdot E=0$, so: \begin{align*} (\pi^*C)\cdot D'&=(\pi^*C)\cdot(\pi^*(\pi_*D)-mE)\\ &=(\pi^*C)\cdot (\pi^*(\pi_*D))\\ &=C\cdot (\pi_*D). \end{align*}
(Here we've used the fact that $(\pi^*D_1)\cdot (\pi^*D_2)=D_1\cdot D_2$ for any divisors $D_1,D_2$ in $S$). $_\blacksquare$