Now that you have all the Stiefel-Whitney classes written down, the hard part is over. To compute Stiefel-Whitney numbers, recall that these are, by definition, obtained in the following way.
Start with a partition of $4$, that is, a sum of a bunch of positive numbers which give $4$. Here are all five of the options: $1+1+1+1,\, 1+1+2,\, 1+3,\, 2+2,\,$ and $4$.
For each choice, form the corresponding product of Stiefel-Whitney class \begin{align*} 1+1+1+1 &\leftrightarrow w_1 \cup w_1 \cup w_1 \cup w_1 \\ 1+1+2 &\leftrightarrow w_1\cup w_1\cup w_2\\ 1+3 &\leftrightarrow w_1\cup w_3\\ 2+2 &\leftrightarrow w_2\cup w_2 \\ 4&\leftrightarrow w_4\end{align*}
The point of a partition is that all the cup products on the right land in $H^4(P^2\times P^2;\mathbb{Z}/2)$. Since every manifold has an orientation class mod $2$, we can pair the element on the right with the orientation class and get a number mod $2$ out. These numbers mod $2$ are the Stiefel-Whitney numbers.
By Poincare duality, the orientation class is the dual of the unique element in $H^4(P^2\times P^2,\mathbb{Z}/2)$, that is, it's the dual of $a^2 b^2$. Hence, computing all the Stiefel-Whitney numbers is the same as computing all the above cup products (using the relations $a^3 = b^3 = 0$), and then counting, mod $2$, the number of occurrences of $a^2 b^2$.
Doing this (while supressing the cup product sign) gives \begin{align*} (w_1)^4 &= (a+b)^4 & &= 0\\ (w_1)^2 w_2 &= (a+b)^2(a^2 + b^2 + ab) & &= 0 \\ w_1 w_3 &= (a+b)(ab^2 + a^2 b) & &= 0\\ (w_2)^2 &= (a^2+b^2+ab)^2 & &= a^2 b^2\\ w_4 &= a^2b^2 & &= a^2 b^2.\end{align*}
(Note that the computations are considerable eased by noting we're working mod $2$ so $(a+b)^2 = a^2 + b^2$.)
From this calculation, we see that three of the Stiefel-Whitney numbers are $0$ (mod $2$) while the other two are $1$ (mod $2$).
I provided a solution to this exercise in my note here; I have copied the proof below. Note, this may not be the exact proof Milnor had in mind.
Lemma: Let $L_1$ and $L_2$ be real line bundles over a paracompact space $B$. Then $w_1(L_1\otimes L_2) = w_1(L_1) + w_1(L_2)$.
Proof:
Let $\pi_i : \mathbb{RP}^{\infty}\times\mathbb{RP}^{\infty}$ denote projection onto the $i^{\text{th}}$ factor and let $\mu : \mathbb{RP}^{\infty}\times\mathbb{RP}^{\infty} \to \mathbb{RP}^{\infty}$ be a classifying map for $\pi_1^*\gamma\otimes\pi_2^*\gamma$. By the Künneth theorem, $\pi_1^*w_1(\gamma)$ and $\pi_2^*w_1(\gamma)$ form a basis for $H^1(\mathbb{RP}^{\infty}\times\mathbb{RP}^{\infty}; \mathbb{Z}_2)$, so $w_1(\pi_1^*\gamma\otimes\pi_2^*\gamma) = a\pi_1^*w_1(\gamma) + b\pi_2^*w_1(\gamma)$ for some $a, b \in \mathbb{Z}_2$.
If $\sigma : \mathbb{RP}^{\infty}\times\mathbb{RP}^{\infty} \to \mathbb{RP}^{\infty}\times\mathbb{RP}^{\infty}$ is the map which interchanges factors, then $\pi_1\circ\sigma = \pi_2$ and $\pi_2\circ\sigma = \pi_1$, so $\sigma^*\mu^*w_1(\gamma) = a\pi_2^*w_1(\gamma) + b\pi_1^*w_1(\gamma)$, but $\sigma\circ\mu$ classifies $\pi_2^*\gamma\otimes\pi_1^*\gamma \cong \pi_1^*\gamma\otimes\pi_2^*\gamma$ so $\sigma\circ\mu$ is homotopic to $\mu$. Therefore
$$a\pi_2^*w_1(\gamma) + b\pi_1^*w_1(\gamma) = (\sigma\circ\mu)^*w_1(\gamma) = \mu^*w_1(\gamma) = a\pi_1^*w_1(\gamma) + b\pi_2^*w_1(\gamma),$$
which implies $a = b$. So either $w_1(\pi_1^*\gamma\otimes\pi_2^*\gamma) = \pi_1^*w_1(\gamma) + \pi_2^*w_1(\gamma)$, or $w_1(\pi_1^*\gamma\otimes\pi_2^*\gamma) = 0$.
Now let $f_i : B \to \mathbb{RP}^{\infty}$ be a classifying map for $L_i$. Then
\begin{align*}
(f_1, f_2)^*(\pi_1^*\gamma\otimes\pi_2^*\gamma) &\cong ((f_1, f_2)^*\pi_1^*\gamma)\otimes((f_1, f_2)^*\pi_2^*\gamma)\\
&\cong (\pi_1\circ(f_1, f_2))^*\gamma\otimes (\pi_2\circ(f_1, f_2)^*)\gamma\\
&\cong f_1^*\gamma\otimes f_2^*\gamma\\
&\cong L_1\otimes L_2.
\end{align*}
As $w_1(L_1\otimes L_2) = w_1((f_1, f_2)^*(\pi_1^*\gamma\otimes\pi_2^*\gamma) = (f_1, f_2)^*w_1(\pi_1^*\gamma\otimes\pi_2^*\gamma)$, if $w_1(\pi_1^*\gamma\otimes\pi_2^*\gamma) = 0$, then $w_1(L_1\otimes L_2) = 0$. This is clearly false, just take $L_1$ to be non-trivial and $L_2$ to be trivial. Therefore $w_1(\pi_1^*\gamma\otimes\pi_2^*\gamma) = \pi_1^*w_1(\gamma) + \pi_2^*w_1(\gamma)$ and so
\begin{align*}
w_1(L_1\otimes L_2) &= (f_1, f_2)^*w_1(\pi_1^*\gamma\otimes\pi_2^*\gamma)\\
&= (f_1, f_2)^*(\pi_1^*w_1(\gamma) + \pi_2^*w_1(\gamma))\\
&= (f_1, f_2)^*\pi_1^*w_1(\gamma) + (f_1, f_2)^*\pi_2^*w_1(\gamma)\\
&= (\pi_1\circ(f_1, f_2))^*w_1(\gamma) + (\pi_2\circ(f_1, f_2))^*w_1(\gamma)\\
&= f_1^*w_1(\gamma) + f_2^*w_1(\gamma)\\
&= w_1(f_1^*\gamma) + w_1(f_2^*\gamma)\\
&= w_1(L_1) + w_1(L_2).
\end{align*}
$$\tag*{$\square$}$$
With this lemma in hand, we can move on to the general case thanks to the splitting principle.
Theorem: Let $E$ and $F$ be real vector bundles over a paracompact space $B$. Let $m = \operatorname{rank} E$ and $n = \operatorname{rank} F$. Then $w(E\otimes F) = p_{m, n}(w_1(E), \dots, w_m(E), w_1(F), \dots, w_n(F))$ where $p_{m, n}$ is the unique polynomial which satisfies
$$p_{m, n}(\sigma_1, \dots, \sigma_m, \tau_1, \dots, \tau_n) = \prod_{i=1}^m\prod_{j=1}^n(1 + x_i + y_j)$$
where $\sigma_k = \sigma_k(x_1, \dots, x_m)$ and $\tau_k = \tau_k(y_1, \dots, y_n)$ are the $k^{\text{th}}$ elementary symmetric polynomials in $m$ and $n$ variables respectively.
Proof: By the splitting principle, there is a paracompact space $Y$ and a map $g : Y \to B$ such that $g^*E \cong \ell_1'\oplus\dots\oplus\ell_m'$ and $g^* : H^*(Y; \mathbb{Z}_2) \to H^*(B; \mathbb{Z}_2)$ is injective. Again by the splitting principle, there is a paracompact space $X$ and a map $f : X \to Y$ such that $f^*g^*F \cong \eta_1\oplus\dots\oplus\eta_n$, and $f^* : H^*(X; \mathbb{Z}_2) \to H^*(Y; \mathbb{Z}_2)$ is injective. Letting $\ell_i = f^*\ell_i'$, we have
$f^*g^*E \cong \ell_1\oplus\dots\oplus\ell_m$. So
$$f^*g^*(E\otimes F) \cong (f^*g^*E)\otimes(f^*g^*F) \cong (\ell_1\oplus\dots\oplus\ell_m)\otimes(\eta_1\oplus\dots\oplus\eta_n) \cong \bigoplus_{i=1}^m\bigoplus_{j=1}^n\ell_i\otimes\eta_j.$$
Therefore,
\begin{align*}
w(f^*g^*(E\otimes F)) &= w\left(\bigoplus_{i=1}^m\bigoplus_{j=1}^n\ell_i\otimes\eta_j\right)\\
&= \prod_{i=1}^m\prod_{j=1}^nw(\ell_i\otimes\eta_j)\\
&= \prod_{i=1}^m\prod_{j=1}^n(1 + w_1(\ell_i\otimes\eta_j))\\
&= \prod_{i=1}^m\prod_{j=1}^n(1 + w_1(\ell_i) + w_1(\eta_j))\\
&= \prod_{i=1}^m\prod_{j=1}^n(1 + x_i + y_j)
\end{align*}
where the penultimate equality uses the lemma and $x_i := w_1(\ell_i)$, $y_j := w_1(\eta_j)$.
Denote the final expression above by $q(x_1, \dots, x_m, y_1, \dots, y_n)$. Note that $q$ is a polynomial which is symmetric in the $x_i$ and the $y_j$ separately, so by the fundamental theorem of symmetric polynomials, there is a unique polynomial $p_{m,n}$ such that
$$q(x_1, \dots, x_m, y_1, \dots, y_m) = p_{m,n}(\sigma_1, \dots, \sigma_m, \tau_1, \dots, \tau_n).$$
Now note that $\sigma_i(x_1, \dots, x_m) = w_i(\ell_1\oplus\dots\oplus\ell_m) = w_i(f^*g^*E) = f^*g^*w_i(E)$ and likewise $\tau_j(y_1, \dots, y_n) = f^*g^*w_j(F)$, so
\begin{align*}
f^*g^*w(E\otimes F) &= w(f^*g^*(E\otimes F))\\
&= q(x_1, \dots, x_m, y_1, \dots, y_n)\\
&= p_{m,n}(\sigma_1, \dots, \sigma_m, \tau_1, \dots, \tau_n)\\
&= p_{m,n}(f^*g^*w_1(E), \dots, f^*g^*w_m(E), f^*g^*w_1(F), \dots, f^*g^*w_n(F))\\
&= f^*g^*p_{m,n}(w_1(E), \dots, w_m(E), w_1(F), \dots, w_n(F)).
\end{align*}
By the injectivity of $f^*$ and $g^*$, we have $w(E\otimes F) = p_{m,n}(w_1(E), \dots, w_m(E), w_1(F), \dots, w_n(F))$.
$$\tag*{$\square$}$$
Best Answer
As explained in the comments, the assertion
is incorrect; that's not how the naturality axiom works. (The problem was that Milnor--Stasheff's definition of bundle map isn't what you might reasonably expect it to be.)
Anyway, to solve the problem, you just need to write the Cartesian product bundle as $p_1^* \xi \oplus p_2^* \eta$ and then apply the Whitney sum formula.