In dimension 2, if you pull off the half-segment $\{1/2\}\times(0,1/2]$ from the square $(0,1)^2$ you obtain an open set $\Omega$ which does not satisfy the lipschitz condition (nor the cone condition). This kind of "fissure" domains are helpful to test the necessity of the cone condition. More precisely, if $B$ is a small ball around $(1/2,0)$ (say of radius $1/4$), its intersection with $\Omega$ as two connected components (one at the left of the above half-segment, another one on the right). Now you can easily find a smooth function $f\in\mathscr{C}^\infty(\Omega)$ equaling $1$ one on the first connected component of $B\cap \Omega$ and $0$ on the other one, with all derivatives bounded. This already tells you that no $W^{1,1}(\mathbf{R}^2)$ extension of $f$ is possible even though $f$ belongs to all Sobolev spaces $W^{1,p}(\Omega)$ for $p\in[1,\infty]$. 

However, as far as compactness issues are concerned, I believe that a single fissure as in the previous example is not enough to contradict the Rellich-Kondrachov Theorem : $\Omega$ as above is the union of 4 Lipschitz domain (four small squares, in fact even 2 small squares and 1 rectangle would be enough) on which you can invoke each time Rellich-Kondrachov theorem and recover convergence (finite number of sequences implies common extraction).

If this " 4 pieces " argument is correct, this already tells you that the right assumptions on the boundary for the Rellich-Kondrachov are not necessarily the same as the one of the  extension theorems. 

EDIT: Rellich-Kondrachov contains two statements : the embedding in $L^{p^\star}(\Omega)$ and the compact embedding in $L^q(\Omega)$ for $q<p^\star$. In the situation at stake in which $\Omega$ is not regular enough, the whole issue boils down to the first statement because we clearly have compact embedding in $L_{\text{loc}}^q(\Omega)$ (looking at balls contained in $\Omega$ there's no regularity issue) so for the second statement of the Rellich-Kondrachov theorem (compactness), we only need to check uniform integrability in $L^q(\Omega)$ (I focus here on the case when $\Omega$ is bounded) ; obviously this is true once the embedding in $L^{p^\star}(\Omega)$ is checked. 

On the other hand, for a counterexample in which the embedding $W^{1,p}(\Omega)\hookrightarrow L^q(\Omega)$ is false for any $q>p$, you can check the book of Adams and Fournier, Theorem 4.48 (the relevant pathology is the exponential cusp, not the fissure). I would be interested to know if this counterexample can lead to a sequence $(f_n)_n$ bounded in $W^{1,p}(\Omega)$ but not uniformly integrable in $L^p(\Omega)$ (and thus not compact).