Suppose we have two independent zero mean random variables $V$ and $U$. Now let $W=V+U$.
It is not difficult to show that \begin{align} E[ E[V|W]\cdot W] \neq 0 \end{align} The proof goes as follows \begin{align} E[ E[V|W]\cdot W] &= E[ V W ], \text{ by orthogonality } E[ (V-E[V|W]) \cdot W]=0\\ &=E[ V (V+U) ]= E[ V^2]>0 \end{align}
My questions can we show a modified statment \begin{align} E[ E[V|W]\cdot W \cdot g(W)] \neq 0 \end{align} for any $g(W)>0$ and $V$ and $W$ with symmetric distributions.
Assume that both $E[V^2]>0$ and $E[W^2]>0$ and $V$ and $W$ are not degenerate.
No, we can't. For example, let $V$ be $\pm 1$ with equal probabilities, while $U = -3, 0, 3$ with equal probabilities. We get $W =-4, -2, -1, 1, 2, 4$ with equal probabilities and $E[V|W] = -1,1,-1,1,-1,1$ respectively. Thus $$E[E[V|W] W g(W)] = 4 g(-4) - 2 g(-2) + g(-1) + g(1) - 2 g(2) + 4 g(4)$$ and there is no reason for that to be nonzero. For example, you might have $g(\pm 4) = g(\pm 1) = 1$ and $g(\pm 2) = 5/2$.