Let $\Omega \subset \mathbb{R}^n$. The space $W_0^{k,p}(\Omega)$ is defined as the closure of $C_0^{\infty}(\Omega)$ in $W^{k,p}(\Omega)$.
I can't say I fully understand the rationale of this definition. Would it be wrong to just say that $W_0^{k,p}(\Omega)$ is the set of compactly supported $W^{k,p}(\Omega)$ functions?
That's not true. For example, when $n=1$, $\sin x$ is an element in $W^{1,p}_0(0,\pi)$ since it is a $W^{1,p}$-limit of elements in $C^\infty_c((0,\pi))$. However, $\sin x$ does not have compact support $(0,\pi)$. In general, if $\Omega$ has a nice boundary (for example if it's $C^1$), $W^{k,p}_0(\Omega)$ are those in $W^{k,p}(\Omega)$ which vanishes at the boundary. To be precise, there is a bounded trace operator
$$T : W^{k,p}(\Omega) \to L^p(\partial \Omega)$$
so that $Tu = u|_{\partial \Omega)}$ if $u\in W^{k,p}(\Omega) \cap C(\overline\Omega)$. And we have
$$ W^{k,p}_0(\Omega) = \{ f\in W^{k,p}(\Omega): Tu = 0\}.$$
The proof can be found in Evans' PDE for example.