Let $V$ be a finite dimensional vector space over a field $k$. It is known that the symmetric algebra $S = S(V)$ is "dual" to the exterior algebra $E = \bigwedge(V^*)$. Now choose a basis $\{x_1,\ldots, x_n \}$ of $E$ and a dual basis $\{s_1,\ldots, s_n\}$ of $S$.
Let $(M,d)$ be a Differential graded $E$-module. Consider the $S$-module $N = M \otimes_k S$ and the map $\tilde{d}$ defined by
$\tilde{d}(a \otimes s) = d(a) \otimes s + \sum_{i=1}^n x_i a \otimes s_i s$.
The claim is that $(N,\tilde{d})$ is a differential graded module. I have trouble showing that $\tilde{d}^2 = 0$. In fact,
$\tilde{d}^2 (a \otimes s) = d^2a \otimes s + \sum_i x_id(a)\otimes s_is + \sum_{i,j} d(x_ia) \otimes s_ia + x_ix_j a + s_is_j s $
and I do not know how to check that that this is zero besides the obvious terms $d^2(a)$ and $x_i^2$.
What you've written doesn't look quite right - there's some confusion with indices. You should get $$\tilde{d}^2(a\otimes s)=\sum_i x_i d(m) \otimes s_i s + \sum_i d(x_ia)\otimes s_is + \sum_{ij} x_i x_j a \otimes s_is_js$$ where I've missed off the term with $d^2$ since $d^2=0$.
The term $\sum_i x_i d(a)\otimes s_is$ cancels with $\sum_i d(x_i a)\otimes s_is$ as $d$ is a differential.
The last sum is zero: the $i=j$ terms are zero as $x_i^2=0$ and $$ x_ix_j a \otimes s_i s_j s = - x_j x_i a \otimes s_j s_i s$$ as $E$ is anticommutative and $S$ is commutative, so terms with $i\neq j$ in the last sum come in pairs summing to zero.