type-definable over $M$ equivalence relation on $G^*$ containing the relation $F_\Delta$. We easily see that all the relations ${F_0}_\Delta$, $F_\Delta$ and ${\bar F}_\Delta$ are coarser than the relation of having the same complete $\Delta$-type over $M$, so the last two are bounded equivalence relations.

As usual, in the definable context (i.e. when $\Delta$ consists of all formulas of the appropriate form), we skip the index $\Delta$ and write $F_0$, $F$ and $\bar F$.

The relations ${F_0}_\Delta$, $F_\Delta$ and ${\bar F}_\Delta$ are all invariant under the action of $G$, so $(G,G^*/{\bar F}_\Delta,e/{\bar F}_\Delta)$ is a $G$-ambit.

First, we check that ${F_0}_\Delta$ is invariant under $G$. Consider any $g \in G$ and $\alpha,\beta \in G^*$ such that ${F_0}_\Delta(\alpha,\beta)$. Take $a,b,a_1,b_1$ from the definition of ${F_0}_\Delta$ witnessing that ${F_0}_\Delta(\alpha,\beta)$ holds. Since $g \in G \subseteq M$, one easily checks that $ga,b,ga_1,b_1$ witness that ${F_0}_\Delta(g\alpha,g\beta)$ holds.

Since ${F_0}_\Delta$ is invariant under $G$, so is $F_\Delta$. For any $g \in G$, the equivalence relation $g{\bar F}_\Delta$ is type-definable over $M$ and contains $gF_\Delta=F_\Delta$, so $\bar F_\Delta \subseteq  g \bar F_\Delta$. This implies that ${\bar F}_\Delta$ is invariant under $G$, which yields the natural action of $G$ on $G^*/{\bar F}_\Delta$ given by $g(h/{\bar F}_\Delta):= (gh)/{\bar F}_\Delta$. 