I don't think it is really clear from that Wikipedia page, but it is a bit clearer in May's book.
You need to consider the construction of the Stiefel-Whitney classes on page 197 of May's book.
i.e. There is a map $\omega^\ast:H^*(BO(n);\mathbb{Z}_2) \to H^*((\mathbb{R} P^\infty;\mathbb{Z}_2)^{\Sigma_n}$ which is an isomorphism and we define the Stiefel-Whitney classes by letting $w_i$ be the unique element such that $\omega^*(w_i)=\sigma_i$, where $\sigma_i$ which is the $i$-th elementary symmetric polynomial (which makes sense since $H^*(BO(n);\mathbb{Z}_2)$ is a $\mathbb{Z}_2$-polynomial ring on the elementary symmetric polynomials), $w_0=1$ and $w_i=0$ for $i > n$.
So we need to know what the Steenrod operations on $\mathbb{R} P^\infty$ are. Since this is a ring generated by a one-dimensional class, this is not too hard to do:
Claim: For $u \in H^1(\mathbb{R} P^\infty;\mathbb{Z}_2)$, $\text{Sq}^i(u^j) = \binom{j}{i}u^{j+1}$
Proof: The total Steenrod squaring operation is a ring homomorphism and $\text{Sq}(u) = \text{Sq}^0(u) + \text{Sq}^1(u) = u + u^2$, hence $\text{Sq}(u^j) = (u+u^2)^j$ and then you can get the result from the binomial formula.
Then bump this up using the Cartan formula, and fingers crossed, you get something that looks like the Wikipedia page.
The other approach is to open up your (well worn, read, and studied) copy of Milnor-Staffesh. In Chapter 8 they give another definition of the Stiefel-Whitney classes. See page 91, where they use the Thom isomorphism to define the Stiefel-Whitney classes.
Without explaining all the terms:
In other words $w_i(\xi)$ is the unique cohomology classes in $H^i(B;\mathbb{Z}_2)$ such that $\phi(w_i(\xi)) = \pi^*w_i(\xi) \smile u$ is equal to $\text{Sq}^i\phi(1) = \text{Sq}^i(u)$.
(If you want to read more consult Milnor-Stasheff or read Akhil Matthew's blog post).
Then problem 8.1 asks you to derive the exact formula given on the Wikipedia page (it should go something like - it is true for line bundles by the claim above. Then if it is true for $\xi$ it is true for $\xi \oplus L$ where $L$=line bundle. Then use the splitting principle)