Let $S\mathbb{Z}_2^n := (\epsilon_1,...,\epsilon_n) : \sum_i \epsilon_i = 0 \mod 2$. So we have embedding $S\mathbb{Z}_2^n \to \mathbb{Z}_2^n$ and, hence, embedding $BS\mathbb{Z}_2^n \to B\mathbb{Z}_2^n$. So, we have morphism between cohomology: $H^*(B\mathbb{Z}_2^n,\mathbb{Z}_2) \to H^*(BS\mathbb{Z}_2^n,\mathbb{Z}_2)$.
I understand, that $H^*(B\mathbb{Z}_2^n,\mathbb{Z}_2) = \mathbb{Z}_2[t_1,...,t_n]$ and, because $S\mathbb{Z}_2^n = \mathbb{Z}_2^{n-1}$ we have $H^*(BS\mathbb{Z}_2^n,\mathbb{Z}_2) = \mathbb{Z}_2[p_1,...,p_{n-1}]$. But I dont understand how acts morphism: $\mathbb{Z}_2[t_1,...,t_n] \to \mathbb{Z}_2[p_1,...,p_{n-1}]$ induced from cohomology. Hope for your help!
You may not have realized it, but you've made a sleight-of-hand in setting up your question. Namely, you write "because $S\mathbb{Z}_2^n = \mathbb{Z}_2^{n-1}$ we have $H^*(BS\mathbb{Z}_2^n, \mathbb{Z}_2) = \mathbb{Z}_2[p_1, \ldots, p_{n-1}]$''. However, the isomorphism $S\mathbb{Z}_2^n \cong \mathbb{Z}_2^{n-1}$ is not canonically determined, so neither are the coordinates $p_1, \ldots, p_{n-1}$. Once you have specified the isomorphism $S\mathbb{Z}_2^n \cong \mathbb{Z}_2^{n-1}$, then the answer should become obvious.
It suffices to understand the induced map in cohomology just in degree 1. Consider the elements $t_i \in H^1(B\mathbb{Z}_2^n, \mathbb{Z}_2)$. We may interpret $t_i$ as the group homomorphism $\mathbb{Z}_2^n \to \mathbb{Z}_2$ that is the projection to the $i$-th component. Now, the induced map $H^1(B\mathbb{Z}_2^n, \mathbb{Z}_2) \to H^1(BS\mathbb{Z}_2^n, \mathbb{Z}_2)$ is given by restriction of a group homomorphism from $\mathbb{Z}_2^n$ to the subgroup $S\mathbb{Z}_2^n$.
For example, if we think of $(\mathbb{Z}_2)^n$ as a $\mathbb{Z}_2$-vector space with coordinates $e_1, e_2, \ldots, e_n$, and realize $S(\mathbb{Z}_2)^n$ as the vector subspace spanned by $e_1 + e_n, e_2 + e_n, \ldots, e_{n-1} + e_n$, then we can identify $p_i: S\mathbb{Z}_2 \to \mathbb{Z}_2$ as the element dual to $e_i + e_n$. For $i < n$ and all $j$, we have $$t_i(e_j + e_n) = \delta_{ij} = p_i(e_j + e_n),$$ so we have $t_i \mapsto p_i$. Similarly, when $i = n$, we have $$t_n(e_j + e_n) = 1 = \sum_{i=1}^{n-1} p_i(e_j + e_n),$$ so $t_n \mapsto \sum_{i=1}^{n-1} p_i$. Of course, this is just the linear dual of the inclusion map $S\mathbb{Z}_2^n \to \mathbb{Z}_2^n$, which we can only write in coordinates once a basis for $S\mathbb{Z}_2^n$ has been chosen.