Suppose $r, n, n \in \mathbb{N}$ and we have an axial map $f \colon \mathbb{R}P^s \times \mathbb{R}P^r \to \mathbb{R}P^n$, that means that there are basepoints, such that for the restricted maps $* \times \mathbb{R}P^r \to\mathbb{R}P^n$ and $\mathbb{R}P^s \times * \to \mathbb{R}P^n$, the generator in $H^1(\mathbb{R}P^n; \mathbb{F}_2) \cong \mathbb{F}_2$ gets send to the basepoints in $H^1(\mathbb{R}P^r; \mathbb{F}_2)$ and $H^1(\mathbb{R}P^s; \mathbb{F}_2)$ (I guess that this means exactly that $f$ induces isomorphisms in cohomology?).
I want to prove, that then there is a homotopy-commutative diagram $\require{AMScd}$ \begin{CD} \mathbb{R}P^r \times \mathbb{R}P^s @>{j \times j}>> \mathbb{R}P^{\infty} \times \mathbb{R}P^{\infty}\\ @VV{f}V @VV{\mu}V\\ \mathbb{R}P^n @>{j}>> \mathbb{R}P^{\infty}, \end{CD} where $j$ is the respective inclusion, and $\mu \colon \mathbb{R}P^{\infty} \times \mathbb{R}P^{\infty} \to \mathbb{R}P^{\infty}$ is the map representing tensor products of line bundles.
I have the following hint: Since $f$ is axial, the induced map $f^* \colon H^1(\mathbb{R}P^n; \mathbb{F}_2) \to H^1(\mathbb{R}P^s \times \mathbb{R}P^r; \mathbb{F}_2)$ sends the generator $x$ to $x \otimes 1 + 1 \otimes x$, and one should use that $\mathbb{R}P^{\infty}$ represents $H^1(-, \mathbb{F}_2)$, i.e. for good spaces $X$, it is $H^1(X; \mathbb{F}_2) \cong [X, \mathbb{R}P^{\infty}]$.
But I dont now neither how to prove the hint, nor how to get from the hint to the final statement. My idea for the hint is, that $\mathbb{R}P^s \times \mathbb{R}P^r \cong (* \times \mathbb{R}P^r ) \times (\mathbb{R}P^s \times *)$ and then using the Künneth formula, but essentially, I dont know how to get from the restricted pullbacks to the pullback $f^*$ of $f$.
You have two maps $\mathbb{R}P^r \times \mathbb{R}P^s \to \mathbb{R}P^\infty$ from tracing out the two paths in the square diagram, and you want to show that they are homotopic. As you point out, since $\mathbb{R}P^\infty$ is a $K(\mathbb{F}_2, 1)$, these maps are classified up to homotopy by a cohomology class in $H^1(\mathbb{R}P^r \times \mathbb{R}P^s; \mathbb{F}_2)$, so it suffices to show that the classes associated to the two maps are the same. Below, we shall write $H^1(-)$ for $H^1(-; \mathbb{F}_2)$ for convenience.
Consider first the composite $\mathbb{R}P^r \times \mathbb{R}P^s \xrightarrow{f} \mathbb{R}P^n \xrightarrow{j} \mathbb{R}P^\infty$. Let $x$ be the generator of $H^1(\mathbb{R}P^\infty)$; since $j$ induces an isomorphism on $H^1$, we abuse notation and write $x$ for the generator of $H^1(\mathbb{R}P^n)$ (and $H^1(\mathbb{R}P^s)$, etc.) as well.
To prove this, first note a general element of $H^1(\mathbb{R}P^r \times \mathbb{R}P^s)$ has the form $a (1 \otimes x) + b (x \otimes 1)$ for some $a, b \in \mathbb{F}_2$; this follows from the Kunneth formula. So write $f^* x = a (1 \otimes x) + b (x \otimes 1)$, and we want to show that $a = b = 1$. First, consider the inclusion $i_1: \mathbb{R}P^r \to \mathbb{R}P^r \times \mathbb{R}P^s$ as the first factor. Since $f$ is axial, the composite $f \circ i_1$ induces an isomorphism in $H^1$, so $i_1^* f^*(x) = x$, i.e., $a i_1^*(1 \otimes x) + b i_1^*(x \otimes 1) = x$. Now, $i_1^*(1 \otimes x) = 0$ and $i_1^*(x \otimes 1) = x$, so this equation becomes $b x = x$, i.e., $b = 1$. Arguing similarly with the other inclusion $i_2: \mathbb{R}P^s \to \mathbb{R}P^r \times \mathbb{R}P^s$ establishes $a = 1$. Thus $f^* x = 1 \otimes x + x \otimes 1$, as wanted.
Returning to the composite $\mathbb{R}P^r \times \mathbb{R}P^s \xrightarrow{f} \mathbb{R}P^n \xrightarrow{j} \mathbb{R}P^\infty$, the previous claim shows that the composite is classified by $f^* j^* x = 1 \otimes x + x \otimes 1 \in H^1(\mathbb{R}P^r \times \mathbb{R}P^s)$.
Now for the other composite $\mathbb{R}P^r \times \mathbb{R}P^s \xrightarrow{j \times j} \mathbb{R}P^\infty \times \mathbb{R}P^\infty \xrightarrow{\mu} \mathbb{R}P^\infty$, note that $\mu$ is also axial (using your definition, this follows from the fact that the formula $w_1(L_1 \otimes L_2) = w_1(L_1) + w_1(L_2)$ for the Stiefel-Whitney classes of the tensor product of line bundles), i.e., $\mu^* x = 1 \otimes x + x \otimes 1$. Therefore, the composite map is classified by $(j \times j)^* \mu^* x$, which is also $1 \otimes x + x \otimes 1$.
Since the two maps around the square are classified by the same element, they are homotopic.