What does this map $\mathbb{R}P^\infty\to\mathbb{C}P^\infty$ induce on cohomology?

First of all, your formula for the cohomology of $\mathbb{R}P^\infty$ is wrong (I think you may have gotten confused with mod $2$ cohomology). The correct formula is $H^*(\mathbb{R}P^\infty)\cong\mathbb{Z}[t]/(2t)$ where $|t|=2$. In any case, the induced map $f^*$ on cohomology is determined by $f^*(y)$, since it must be a ring homomorphism.

Here is one way to compute $f^*(y)$. First, note that the inclusion map $i:\mathbb{R}P^2\to\mathbb{R}P^\infty$ induces an isomorphism on $H^2$, so actually we can just compute $i^*f^*(y)$.

Now let's think about the map $fi:\mathbb{R}P^2\to\mathbb{C}P^\infty$. This map sends $[x,y,z]$ to $[x,y,z,0,0,\dots]$. Note that this is homotopic to the map that sends $[x,y,z]$ to $[x+iy,z,0,0,\dots]$ (first linearly interpolate from $[x,y,z,\dots]$ to $[x+iy,0,z,\dots]$ and then linearly interpolate to $[x+iy,z,0,\dots]$). Let us call this map $jg$, where $j:\mathbb{C}P^1\to\mathbb{C}P^\infty$ is the inclusion and $g:\mathbb{R}P^2\to\mathbb{C}P^1$ sends $[x,y,z]$ to $[x+iy,z]$. Since $j$ induces an isomorphism on $H^2$, it is enough to compute what the map $g$ does on $H^2$.

But the map $g$ is very easy to understand geometrically. This map $g$ is surjective and is injective except that it sends all points $[x,y,0]$ to $[1,0]$. This means that $g$ is just the quotient map $\mathbb{R}P^2\to S^2\cong\mathbb{C}P^1$ that collapses the $1$-skeleton of $\mathbb{R}P^2$ to a point. In particular, $g$ is cellular, and we can easily compute via cellular cohomology that $g^*:H^2(\mathbb{C}P^1)\to H^2(\mathbb{R}P^2)$ is nontrivial, and more specifically maps $j^*(y)$ to $i^*(t)$. (Here $H^2(\mathbb{C}P^1)\cong\mathbb{Z}$ is generated by $j^*(y)$ and $H^2(\mathbb{R}P^2)\cong\mathbb{Z}/2$ is generated by $i^*(t)$.)

Now we can unwind all the isomorphisms from above to find $f^*(y)$. We have $g^*j^*(y)=i^*(t)$. But $jg$ is homotopic to $fi$, so $i^*f^*(y)=g^*j^*(y)=i^*(t)$. Since $i^*$ is an isomorphism, this implies $f^*(y)=t$.

(Actually, in this case, since the only options are $f^*(y)=0$ and $f^*(y)=t$, it is enough to observe that since $g^*$ was nontrivial, $f^*$ must be nontrivial as well, so the only possibility is $f^*(y)=t$.)