As you say, we define $w_n = \sum \text{Sq}^i \nu_{n - i}$, where $\nu_{n-i}$ is the Wu class, the class such that $\nu_{n-i} \cup c = \text{Sq}^{n-i} c$ for $c \in H^{i}$. So as a corollary we have $\text{Sq}^i \nu_{n - i} = \nu_i \cup \nu_{n-i}$. 

Because $\nu_j$ vanishes for $j > n/2$, the sum over $i$ is only the term $\nu_{n/2}^2$ when $n$ is even, and $0$ when $n$ is odd. As the Euler characteristic of an odd-dimensional closed manifold vanishes, this gives the odd-dimensional case. 

Now when $n = 2k$, using Poincare duality mod 2 we see that $\chi(M) = \text{rk } H^k(M;\Bbb Z/2) \pmod 2.$ So the claim is that $\langle \nu_k^2, [M] \rangle = \text{rk } H^k(M;\Bbb Z/2) \pmod 2.$ 

This is because $\nu_k$ is a characteristic vector for the symmetric bilinear cup-product form on $H^k(M;\Bbb Z/2)$; in fact, for any 'characteristic vector' $y$ for a nondegenerate symmetric bilinear form over a $\Bbb Z/2$-vector space $V$, meaning that $y \cdot x = x^2$ for all $x$, we have $\text{rk } V = y^2 \pmod 2$. 

The most obvious way for me to see this is to classify nondegenerate symmetric bilinear forms over $\Bbb Z/2$ vector spaces: they are all sums of copies of $\begin{pmatrix}1\end{pmatrix}$ and $\begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}$, for which the respective characteristic vectors are $(1)$ and $(0,0)$.