Suppose $R$ is a finite commutative local ring of characteristic $2^n$ with unique maximal $M$ and its residue field $k$ (of characteristic $2$). Consider the canonical map $\pi : R \rightarrow k$ by $\pi (x) = \bar{x} $.
Is it true that for all $\alpha \in R, \alpha \in \{ x^2 + x : x \in R \}$ if and only if $\bar{\alpha} \in \{ \bar{y}^2 + \bar{y} : \bar{y} \in k \}$?
Yes. First, note that $ X^2 + X = Y^2 + Y $ entails $ (X-Y)(X+Y) = Y-X $ or
$$ (X-Y)(X+Y+1) = 0 $$
Since at least one of $ X - Y $ and $ X + Y + 1 $ is a unit, it follows that either $ X = Y $ or $ X = -Y - 1 $. These cannot be equal, as then the characteristic $ 2^n $ would not annihilate $ Y $. It follows that the map $ X \to X + X^2 $ defined on $ R $, and likewise on the residue field $ k = R/M $, is two-to-one. Let $ m $ denote the cardinality of $ M $, and $ r $ the cardinality of $ R $. The preimage of a subset $ S $ of $ k $ under the surjective homomorphism $ \pi $ has cardinality $ |S| m $. For the sets $ S, S' $ defined by
$$ S = \{ \alpha \in k : \alpha = x + x^2, x \in k \} $$ $$ S' = \{ \beta \in R : \beta = x + x^2, x \in R \} $$
clearly $ S' \subset \pi^{-1}(S) $. However, we have
$$ |\pi^{-1}(S)| = m|S| = \frac{m|k|}{2} = \frac{mr}{2m} = \frac{r}{2} = |S'| $$
It follows that $ S' = \pi^{-1}(S) $, which is the desired result.