Here is a related example from:

P. Hajlasz, Sobolev mappings: Lipschitz density is not a bi-Lipschitz invariant of the target. Geom. Funct. Anal. 17 (2007), 435-467.



***Theorem.*** *There is a Lipschitz function 
$\varphi\in {\rm Lip}\, (\mathbb{R}^2)$
with compact support such that the bounded operator
$\Phi:W^{1,p}([0,1],\mathbb{R}^2)\to W^{1,p}([0,1])$ defined as
composition $\Phi(u)=\varphi\circ u$ is not continuous for any $1\leq
p<\infty$.*

The operator is bounded in the sense that the Sobolev norm of $\varphi\circ u$ is bounded by constant times that of $u$. However, the operator is not continuous as a mapping between Banach spaces $W^{1,p}([0,1],\mathbb{R}^2)$ and $W^{1,p}([0,1])$.

However as was proved in

M. Marcus,V. J. Mizel, Every superposition operator mapping one Sobolev space
into another is continuous. J. Funct. Anal. 33 (1979), 217-229,

the composition operator in the case in which $\varphi$ is a Lipschitz function on $\mathbb{R}$ is continuous.

Another reference for understanding the chain rule for Sobolev and BV functions is:

Ambrosio, L.; Dal Maso, G. A general chain rule for distributional derivatives. Proc. Amer. Math. Soc. 108 (1990), no. 3, 691-702.