Calculate the Wu class from the Stiefel-Whitney class

First of all, your definition of Wu class is incorrect. If $X$ is a closed connected $n$-manifold, there is a unique class $\nu_k \in H^k(X; \mathbb{Z}_2)$ such that for any $x \in H^{n-k}(X; \mathbb{Z}_2)$, $\operatorname{Sq}^k(x) = \nu_k\cup x$. We call $\nu_k$ the $k^{\text{th}}$ Wu class. If $X$ is also smooth, then the Stiefel-Whitney classes of the tangent bundle of $X$ are related to Steenrod squares and Wu classes by the formula

$$w_i = \sum_{k = 0}^i\operatorname{Sq}^k(\nu_{i-k}).$$

Note $\operatorname{Sq}^k(\nu_{i-k})$ is not simply $\nu_k\cup\nu_{i-k}$ unless $i = n$. So we have

\begin{align*} w_1 &= \operatorname{Sq}^0(\nu_1) = \nu_1\\ w_2 &= \operatorname{Sq}^0(\nu_2) + \operatorname{Sq}^1(\nu_1) = \nu_2 + \nu_1\cup\nu_1\\ w_3 &= \operatorname{Sq}^0(\nu_3) + \operatorname{Sq}^1(\nu_2) = \nu_3 + \operatorname{Sq}^1(\nu_2) \end{align*}

It follows that $\nu_1 = w_1$ and $\nu_2 = w_2 + w_1\cup w_1$. However, at this stage we can only deduce $\nu_3 = w_3 + \operatorname{Sq}^1(\nu_2)$. In order to determine $\nu_3$ in terms of Stiefel-Whitney classes, we need to compute $\operatorname{Sq}^1(\nu_2)$. First note that

\begin{align*} \operatorname{Sq}^1(\nu_2) &= \operatorname{Sq}^1(w_2 + w_1\cup w_1)\\ &= \operatorname{Sq}^1(w_2) + \operatorname{Sq}^1(w_1\cup w_1)\\ &= \operatorname{Sq}^1(w_2) + \operatorname{Sq}^0(w_1)\cup\operatorname{Sq}^1(w_1) + \operatorname{Sq}^1(w_1)\cup\operatorname{Sq}^0(w_1) && \text{(by Cartan's formula)}\\ &= \operatorname{Sq}^1(w_2) \end{align*}

so $\nu_3 = w_3 + \operatorname{Sq}^1(w_2)$. To compute Steenrod squares of Stiefel-Whitney classes, we use Wu's formula

$$\operatorname{Sq}^i(w_j) = \sum_{t=0}^k\binom{j-i+t-1}{t}w_{i-t}\cup w_{j+t}.$$

In this case, we see that

$$\operatorname{Sq}^1(w_2) = \binom{0}{0}w_1\cup w_2 + \binom{1}{1}w_0\cup w_3 = w_1\cup w_2 + w_3.$$

Therefore, $\nu_3 = w_3 + \operatorname{Sq}^1(w_2) = w_3 + w_1\cup w_2 + w_3 = w_1\cup w_2$. Suppressing the cup symbol, this agrees with the identity given on nLab.

See this note for more details, as well as the computations for $\nu_4$ and $\nu_5$.