Assume that $\tilde X \to X$ and $\tilde Y\to Y$ are double covers. Then $\mathbb{Z}_2$ acts freely on $\tilde X$ by deck transformations, and likewise it acts freely on $\tilde Y$ by deck transformations. Let $\mathbb{Z}_2$ act on $\tilde X\times \tilde Y$ by the diagonal action. Is there an equivariant Kunneth formula relating the equivariant cohomologies $H^*_{\mathbb{Z}_2}(\tilde X;\mathbb{Z}_2)$, $H^*_{\mathbb{Z}_2}(\tilde Y;\mathbb{Z}_2)$, and $H^*_{\mathbb{Z}_2}(\tilde X\times \tilde Y;\mathbb{Z}_2)$?
Edit: There seems to be an equivariant Kunneth formula (see for example Theorem 61 here) which states that, in nice situations, we have $$ H^*_G(M\times N)\cong H^*_G(M)\otimes_{H^*(BG)}H^*_G(N), $$ where $G$ acts on $M\times N$ by the diagonal action. However, this doesn't seem to apply in the above setting. For example, taking both covers to be $S^2\to \mathbb{R}\text{P}^2$ we have $$ H^*_{\mathbb{Z}_2}(S^2;\mathbb{Z}_2)\otimes_{\mathbb{Z}_2[\lambda]} H^*_{\mathbb{Z}_2}(S^2;\mathbb{Z}_2) \cong\mathbb{Z}_2[\lambda]/(\lambda^3). $$ However, computing $H^*_{\mathbb{Z}_2}(S^2\times S^2;\mathbb{Z}_2)$ from a cell structure shows that $H^2_{\mathbb{Z}_2}(S^2\times S^2;\mathbb{Z}_2) \cong \mathbb{Z}_2\oplus\mathbb{Z}_2$. Perhaps it is still true that the tensor product injects into the cohomology of the product?