Map induced in mod $2$ cohomology of a projection $S^n \to S^n/\mathbb{Z}_2$ Consider the involution $\varphi_i \colon S^n \to S^n$ given by $(x_0, \ldots, x_n) \mapsto (x_0, \ldots, x_{i-1}, -x_i, \ldots, -x_n)$, where $0\leq i\leq n$.
Let $f_i \colon S^n \to S^n/\mathbb{Z}_2$ be the quotient map corresponding to the involution $\varphi_i$. I known that if $i=0$, then $S^n/\mathbb{Z}_2=\mathbb{R}P^n$, and $f_i^* \colon H^*(\mathbb{R}P^n; \mathbb{Z}_2) \to H^*(S^n;\mathbb{Z}_2)$ is zero. This is nicely explained e.g. in Characteristic Classes by Milnor and Stasheff.
The other extreme, $i=n$, is easy: in this case $S^n/\mathbb{Z}_2$ is the $n$-disk, hence contractible, and $f_n^*$ is also zero.  
My question is: can something be said about the behaviour of $f_i^*$ on mod $2$ cohomology for $0<i<n$?
 A: For arbitrary $i$ the $i$-th factor $S^n/\mathbb Z_2$ will be homeomorphic to the join $S^{i-1}\star\mathbb RP^{n-i}$, and action the covering map on the $n$-th homology will be actually zero because on the second component of the join the map is $2$-sheet.
A: The map is always trivial mod 2.
Let me change notation a little bit.  I'm going to write the coordinates of $S^{n+m}$ as $(x_1,...,x_n, y_0,..y_m)$ with involution given by negating the $y$ coordinates.
I'm going to view $S^{n+m}$ as the $n$-fold suspension of $S^m$, $S^{n+m} = \Sigma^n S^m$, where the $S^m$ is given by the $y$ coordinates, and the coordinates on the suspension are given by the $x$ coordinates.
Claim 1:  With this identification, $S^n/\mathbb{Z}_2$ is nothing but the $n$-fold suspension of $\mathbb{R}P^m$.
Proof:  Consider the natural projection map $\pi:S^{n+m} = \Sigma^n S^m\rightarrow \Sigma^n \mathbb{R}P^m$ defined by $$\pi(x_1,...,x_n,y_0,...,y_m) = (x_1,...,x_n, [y_0,...y_m]).$$
This function is obviously continuous and surjective.  Note that this function is invariant under the $\mathbb{Z}_2$ action you are considering:  $\pi(x_1,...,x_n, y_0,...y_m) = \pi(x_1,...,x_n, -y_0,...,-y_m)$.  This implies that $\pi$ descends to a continuous, surjective map $\overline{\pi}:S^n/\mathbb{Z}_2\rightarrow \Sigma^n\mathbb{R}P^n$.
The kicker is that $\overline{\pi}$ is also injective.  So, $\overline{\pi}$ is a continuous bijection between compact Hausdorff spaces, so is a homeomorphism.
Claim 2:  With this identification, $f = \Sigma^n p$ where $p$ is the canonical projection $p:S^m\rightarrow \mathbb{R}P^m$.
Proof:  Omitted
Now, since $p$ is trivial on mod 2 cohomology (as you've already noted), using naturality of suspensions, it follows that $f$ is as well.
