Suppose $R_2 > R_1 > 0$ and consider $\Omega \subset \mathbb{R}^2$ such that
$$ \Omega:=\left(B_{R_2}(0)\setminus B_{R_1}(0)\right)\setminus L $$ where $L:= \{(x,y)\in\mathbb{R}^2\,|\,x < 0, y = 0\}$.
Is this domain Lipschitz? I have little background in this area, but having looked at the definition of a Lipschitz domain here,it seems to me that it is isn't, since e.g. if we take $p = (-R',0) \in \partial\Omega$ where $R' = \frac{R_1+R_2}{2}$ and consider any possible $C$ (as in the definition given there) we cannot satisfy what is being said about $\Omega\cap C$.
In the end I would like to apply Poincaré–Wirtinger inequality which is defined here for $p=2$ as $$ \|u-u_{\Omega}\|_{L^2(\Omega)} \leq C \|\nabla u\|_{L^2(\Omega)}, $$ where $u \in W^{1,2}(\Omega)$. In the setup I am working with I specifically exclude $L$ so across it the function is allowed to jump (which would exclude it from the space if $\Omega$ included $L$). Am I correct in thinking that thus the inequality applies? My hand-wavy intuition is that in fact for any $x\in L \cap\left(B_{R_2}(0)\setminus B_{R_1}(0)\right)$ we define 'traces' from above ($x^+$) and below ($x^-$) as something like $$ u(x^\pm) = \lim_{R\to 0^\pm}u(x_1,x_2+R) $$ and thus in some sense these are 'separate' boundaries when looking from above and from below. Sadly I have next to no knowledge of Measure Theory and would appreciate if somebody could clear my doubts. Thank you!
First of all, you are right: $\Omega$ is not a Lipschitz domain. However, a Poincaré inequality for $W^{k,p}(\Omega)$-functions does hold since $\Omega$ satisfies the cone property. For more details you can look in the book "Nonlinear Analysis" by Gasinski and Papageorgiou.