The following is from Lecture notes of Professor Farrell. A hopefully working link is here(page 95):
https://www.dropbox.com/s/80n4wd6xctpe6yr/Characteristic%20Classes%20%28Sparkie%20%E7%9A%84%E5%86%B2%E7%AA%81%E5%89%AF%E6%9C%AC%202014-02-19%29.pdf
We now reach proposition 7. Let $E$ be a complex vector bundle, then the mod 2 reduction of the total Chern class of $E$ is the total Stiefel-Whitney class of $E$. Since there is no Chern classes in odd dimensions, we have there is no Stiefel-Whitney classes in odd dimension as well.
$\textbf{Proof}$
Suppose $E$ is a line bundle. Then we know $w_{0}(L)=1, w_{1}(L)=0,w_{2}(L)=e_{\mathbb{Z}_{2}}(L)=\phi(e(L))$. On the other hand we know $c_{0}(L)=1, c_{1}(L)=e(L)$. So this verified it for line bundles.
For sum of line bundles we have
$$
E=\oplus^{n}_{i=1}L_{i}
$$
So the total Chern class is
$$
c(E)=c(L_{1}) \cup \cdots \cup c(L_{n})\mapsto_{\phi}\omega(E)=\omega(L_{1}) \cup \cdots \cup \omega(L_{n})
$$
We are now going to use splitting principle. But there is a subtle point. We now discuss it.
By the splitting principle there exist
$$
f:\mathcal{B}\rightarrow B
$$
such that
$$
f^{*}(E)=\oplus_{i=1}^{n}L_{i}
$$ and
$$
f^{*}:H^{*}(B,R)\rightarrow H^{*}(\mathcal{B},R)
$$
is monic.
Therefore by naturality and the fact $f^{*}$ is monic with respect to $\mathbb{Z}_{2}$ coefficients.
$$
\phi(f^{*}(c(E)))=f^{*}(\phi(c(E)))=f^{*}(\omega(E))
$$
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 $\nu_5$.
Best Answer
For a $\mathbb{Z}$-module (aka abelian group) $V$ with a symmetric, unimodular, bilinear form $\mu\colon V\otimes V \to V$, a characteristic element is an element $c\in V$ such that
$$\forall v\in V\ \mu(v,v) \equiv \mu(c,v)\mod 2 $$
This means you don't want an element with $\mathbb{Z}/2$ coefficients, you want an integral class. The correct statement is that
It's not hard to see this algebraically using Wu classes (for a definition and important properties see for example Manifold Atlas or Milnor-Stasheff). In general, for a closed, oriented $4n$-manifold, a characteristic element of its intersection form is an integral lift of the $2n$-th Wu class $v_{2n}$, essentially by definition.
But Wu's formula $Sq(V) = W$ tells us that for a closed, oriented manifold we have $v_2 = w_2$, so if $M$ if $4$-dimensional then $\alpha \in H^2(M;\mathbb{Z})$ is characteristic iff it is an integral lift of $v_2$ iff it is an integral lift of $w_2$.