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.

In the doubly even-dimensional case $n = 4k$, we have the intersection form to exploit; the fact that $\nu_{2k} \cup c = \text{Sq}^{2k} c = c^2$ means that $\nu_{2k}$ is a characteristic vector for the intersection form, we know that $\sigma(M) = \langle \nu_{2k}^2, [M]\rangle \pmod 2$. The Euler characteristic reduces modulo 2 to $\text{rk } H^{2k}(M) = \sigma(M) \pmod 2$, as desired.

For the singly even case, let $M$ be a manifold of dimension $4k+2$. Then $\chi(M \times M) = \chi(M)^2 = \chi(M) \pmod 2$, while $\nu_{4k+2}(M \times M) = \nu_{2k+1}(M) \otimes \nu_{2k+1}(M)$ (they both satisfy the same defining property). In particular, $$\chi(M) = \langle \nu_{4k+2}(M \times M)^2, [M \times M]\rangle = \langle\nu_{2k+1}(M)^2 \otimes \nu_{2k+1}(M)^2, [M \times M]\rangle.$$ The evaluation against the fundamental class simplifies to $\langle \nu_{2k+1}(M)^2, [M]\rangle^2 = \langle \nu_{2k+1}(M), [M]\rangle \pmod 2.$

This implies that $\chi(M) = \langle\nu_{2k+1}(M)^2, [M]\rangle$ for $M$ singly even-dimensional.

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 $(0,0)$.