In the book Sobolev space by Robert A.Adams, there is a definition as follows.
Given integers $n > 1$ and $m > 0$, let $N = N (n, m) $be the number of multi-indices $\alpha=(\alpha_1,\cdots,\alpha_n)$ such that $|\alpha|\leq m$. For each such multi-index $\alpha$ let $\Omega_\alpha$ be a copy of $\Omega$ in a different copy of $\mathbb{R}^n$ , so that the N domains $\Omega_\alpha$ are de facto mutually disjoint.
What in fact is $\Omega_\alpha$. What does "be a copy of $\Omega$ in a different copy of $\mathbb{R}^n$" mean?
The idea is that you want to embed $$ W^{m,p}(\Omega) \hookrightarrow L^p(\Omega^{(m)}) : u \mapsto (D^{\alpha}u)_{\lvert\alpha\rvert \leq m}$$ for suitable $\Omega^{(m)}$, so you can view $W^{m,p}(\Omega)$ as a subset of a Lebesgue space, and hence deduce properties and separability, uniform convexity, and reflexivity.
The authors define this space $\Omega^{(m)}$ as a disjoint union of $N$-copies of $\Omega$, but I think this is wrong and it should actually be a product. Since you're associating to each $u \in W^{m,p}(\Omega)$ the tuple consisting of $u$ and all of its derivatives up to order $m$, elements of $\Omega^{(m)}$ should correspond to such tuples.
This can be made precise by setting $\Omega^{(m)}$ to be the $N$-fold product $\Omega \times \cdots \times \Omega$, and $\Omega_{\alpha}$ corresponds to copy of $\Omega$ which $D^{\alpha}u$ maps to. So then you can think of $\Omega = \prod_{\lvert \alpha \rvert \leq m} \Omega_{\alpha}$, however there's no actual disjoint copy $\Omega_{\alpha}$ in $\Omega^{(m)}$ unless you fix a point $x_0 \in \Omega$ and identify $\Omega_{\alpha}$ with the subset $\{x_0\} \times \cdots \times \Omega_{\alpha} \times \cdots \times \{x_0\}$.