From the relation $Sq(v)=w$ we get that $v_2=w_2+w_1^2$. More specifically By Thm 11.14 (which is just an extended form of the above equality) in M-S you have $w_1=v_1$ and $w_2=Sq^1(v_1)+v_2=v_1^2+v_2$. Plug them together you obtain my first claim.
Recall we have a nice characterisation of the (second) Wu-class, as $$Sq^{2}(x)=v_2\smile x$$ for any $x \in H^{n-k}(M;\Bbb Z_2)$ (See Milnor Stasheff page 132). Since in our case $n=3$, we get that for any $x\in H^1(M;\Bbb Z_2)$, $0=v_2\smile x$, which gives you $v_2=0$.
To see this last equality, assume $v_2 \neq 0$. We would have $$1=\langle v_2, (v_2)_*\rangle=\langle v_2, [M]\frown x\rangle=0$$ Where $(v_2)_* \in H_2(M;\Bbb Z_2)$ is the dual of $v_2$ and Poincaré Duality (mod 2) tells us that there exists an element $x \in H^1(M; \Bbb Z_2)$ such that $[M]\frown x=(v_2)_*$
Therefore you have $0=w_2+w_1^2$, which proves your claim