I'm trying to figure out what condition concerning $w$ and $v$ would be enough for me to infer that $E(wu\mid v)=0$ given that I already know $E(u\mid v)=0$.
Clearly, $w$ is a constant works: $E(wu\mid v)=wE(u\mid v)=0$. But I'm trying to find a weaker condition. I'm contemplating $w$ and $u$ are independent. I write $$ E(wu\mid v)=E[E(wu\mid w,v)\mid v]=E[wE(u\mid w,v)\mid v]=E[wE(u\mid v)\mid v]=E[w0\mid v]=0. $$ Does it seem correct to you? Can you suggest some other sufficient conditions?
Your condition is correct. This can be generalized a little bit more: if $w$ can be written as $w'f(v)$, where $f\colon\mathbf R\to\mathbf R$ is Borel measurable and $w'$ is independent of $u$, then $\mathbb E[wu\mid v]=0$. Indeed, $$\mathbb E[wu\mid v]=f(v)\mathbb E[w'u\mid v]=0$$ by the reasoning in the opening post.