Let $f:\mathbb{R} \to \mathbb{R}$ be a peicewise lipschitz function, eg. $f(x) = \chi_{(0,1)}(x)$.
Let $\Omega$ be a bounded domain in $\mathbb{R}^n$ and let $u \in H^1(\Omega)$. Is the chain rule possible: $(f(u))' = f'(u)u'$ a.e. or something like that?
I have only seen results for $f$ Lipschitz.
As commenters said, taking composition with a discontinuous function like $\chi_{(0,1)}$ takes the function out of all Sobolev classes. It no longer has a weak derivative. For most practical purposes, this composition is useless: we threw out the most valuable information about $u$.
But since you asked about derivative a.e., I'll give an example to show that the composition can fail to have that one too. Let $f=\chi_{(0,1)}$ and $u(x)=\operatorname{dist}(x,C)$ where $C\in [0,1]$ is a Cantor-type set of positive measure. Then $u$ is Lipschitz, therefore in $H^1$ on $(0,1)$. But $f\circ u = 1-\chi_C$, which is discontinuous at every point of $C$. Hence, the composition is not differentiable a.e.