This is Lemma 4.2 from Algebraic Groups and Number Theory by Vladmir Platonov and Andrei Rapinchuk. It is part of a series of arguments for the calculation of a Siegel domain for any reductive group which reduces to that of $\operatorname{GL}_n(\mathbb R)$.
The proof is elementary, but I have tried for awhile to figure it out and have not gotten it.
One should start with an $h \in H$ and write it as $s \gamma$, for $s \in \Sigma$ and $\gamma \in \Gamma$. Then we know that $x.h = x$, or $x.s = x.\gamma^{-1}$. I am at a loss of how to make use of the hypothesis $xa \Sigma \cap x \Gamma = \{xb_1, ... , xb_r\}$.
I see now. The key is to begin with an $h \in H$ and write $a^{-1}h = s \gamma$ for some $s \in \Sigma$ and $\gamma \in \Gamma$. Then $(x.a).a^{-1}h = x.h = x$ and hence $x.as \gamma = x$. Then we can use the hypothesis $xa \Sigma \cap x \Gamma = \{xb_1, ... , xb_r\}$, writing
$$x.as = x.\gamma^{-1} = x.b_i$$
for some $i$. This implies first of all that $\gamma b_i \in \Gamma \cap H$, and second that $h_1 := asb_i^{-1} \in H$, with
$$h = as \gamma = (asb_i^{-1})(b_i \gamma)$$
$$asb_i^{-1} \in a \Sigma b_i^{-1} \cap H \subset \Omega $$
$$b_i \gamma \in \Gamma \cap H.$$
This completes the first assertion of the lemma.
The second assertion assumes that there is a subgroup $\Gamma \subset D \subset G$ for which $\Sigma^{-1} \Sigma \cap g \Gamma h$ is finite for any $g, h \in D$, and we want to prove that $\Omega^{-1} \Omega \cap g(\Gamma \cap H) h$ is also finite for any $g, h \in D \cap H$.
Each $b_i \in \Gamma$ lies in $D$, and hence $\Sigma^{-1} \Sigma \cap (b_i^{-1}g \Gamma \cap H) hb_j)$ is a finite set by our hypothesis. Hence
$$b_i \Sigma^{-1}\Sigma b_j^{-1} \cap g (\Gamma \cap H) h = (a \Sigma b_i)^{-1} (a \Sigma b_j) \cap g (\Gamma \cap H)h$$
is finite, and so is $\Omega^{-1} \Omega \cap g (\Gamma \cap H)h$, as the union of all these sets over all $i, j$.