Implications of mean independence Let $U$ be a random variable with mean $0$. Take other two random variables $X,Y$. Assume
$$
(1)\quad E(U|X,Y)=0.
$$
I believe (1) implies
$$
E(U\cdot X)=E(U \cdot Y)=0.
$$
Does (1) imply
$$
E(U \cdot f(X,Y))=0\quad ?
$$
where $f$ is any function of $X,Y$?
For instance, does (1) imply
$$
E(U \cdot X \cdot Y)=0
$$

Yes.
If $\mathcal A$ denotes a $\sigma$-algebra and $Z$ is a random variable that is $\mathcal A$-measurable then:$$\mathbb E[UZ]=\mathbb E[\mathbb E[U|\mathcal A]Z]$$So if $\mathbb E[U|\mathcal A]=0$ then it follows directly that also $\mathbb E[UZ]=0$.
You can apply that here on the $\sigma$-algebra that is generated by the random variables $X$ and $Y$ and is mostly denoted as $\sigma(X,Y)$.
Beware that $\mathbb E[U|X,Y]$ is actually a notation for $\mathbb E[U|\sigma(X,Y)]$.
Then for Borel measurable $f:\mathbb R^2\to\mathbb R$ we find indeed: $$\mathbb E[Uf(X,Y)]=0$$
You state the condition that $U$ has mean zero, but that is a consequence of your statement $(1)$.
This because:$$\mathbb EU=\mathbb E[\mathbb E[U|X,Y]]$$where the RHS takes value $0$ as a consequence of statement $(1)$.
