Apologies for the long delay in getting back to this. I haven't had time to carefully check the proofs, but I will put my thoughts down now, and if they are correct they should be routine to confirm. Here's what I believe.
First, I believe you may as well replace each $G_i$ with its normal core in $G$, i.e. the intersection of conjugates $\bigcap_{g\in G} gG_ig^{-1}$. The reason this is OK should be that these kinds of questions are usually about closed subsets in the associated topology, and the sequence of $G_i$'s will generate the same topology as their normal cores.
Furthermore, by replacing $G$ by the quotient by the intersection of the $G_i$, we should probably assume that the intersection of the $G_i$ is trivial.
Now, we should consider the completion $\widetilde{G}:=\underleftarrow{\lim} G_i$, a topological group with a natural injection $G\hookrightarrow\widetilde{G}$. I believe (but didn't find time to carefully check) that the sequence "distinguishes the coset $gK$" in your sense if and only if $K$ is closed in the subspace topology.
Assuming this is correct, we can come to your specific questions.
You ask: if $K$ is an infinite-index cyclic subgroup, and the number of orbits of the $G_i$ on $G/K$ tends to infinity, can they still fail to distinguish cosets? Again without checking carefully, I think this is asking if the closure $\overline{K}\cap G$ can be strictly greater than $K$ but still of infinite index in $G$. If this is the question, then the answer is certainly 'yes'. For instance, take a sequence $G_i$ that defines the pro-2 topology on a free group $F_2=\langle a,b\rangle$. Then the closure of $K=\langle a^3\rangle$ is $\langle a\rangle$.
You go on to ask about the case $K=\langle a\rangle$. Here, I believe the answer is 'no', and indeed I believe the answer is 'no' for any maximal cyclic group. It is a basic fact that, as long as the map $G\to\widetilde{G}$ is injective, then centralisers are always closed. This is because if $g\notin C(a)$ then $[a,g]\neq 1$, so under some $q_i:G\to G/G_i$ we see that $q(a)$ does not commute with $q(g)$, whence $g$ is not in the closure of the centraliser $C(a)$. This says that, if $K$ is maximal cyclic, then $K$ is closed in the subspace topology.
Finally, let me just note that your edit imposes a trivial extra condition. Since each $G_i$ has finite index, it coincides with its closure in the profinite completion. Furthermore, the intersection $\bigcap_i G_i$ is an intersection of closed subgroups, hence is itself closed and so equal to its closure.
Anyway, I hope what I wrote here is correct. But even if not, I hope it moves your thoughts along in formulating your question.
The following is my old answer which doesn't answer the question. I've left it in case the references are useful.
Here's a construction using a theorem of Gitik–Rips:
Every free group $F$ is double-coset separable. That is, for any finitely generated subgroups $H,K\leq F$ and any pair of double cosets $HgK\neq Hg'K$, there is a finite quotient $q:F\to Q$ such that $q(Hg'K)\neq q(HgK)$.
(A short proof of this was given by Niblo, who also proved that surface groups have this property, and we now know that many hyperbolic groups have many separable double cosets, thanks to work of Minasyan.)
Now, let $C=\langle g\rangle$ be any infinite cyclic subgroup of a non-abelian free group $F$ and let $K=\langle g^2\rangle$. The number of double cosets $C\backslash F/C$ is always infinite — I can give a proof if needed — but for a simple example, you can take $C=\langle a\rangle$ in $F=\langle a,b\rangle$ and note that the double cosets $Cb^iC$ are all distinct in the abelianisation. Let $g_1,g_2,\dotsc$ be a sequence of double-coset representatives.
By the Gitik–Rips theorem, we may choose a sequence of homomorphisms $q_i$ from $F$ to finite groups such that the double cosets $Cg_1C,\dotsc, Cg_iC$ all have distinct images under $q_i$. Take $G_i=q_i^{-1}(C)$. Then the number of orbits of $G_i$ on $G/C$ is equal to the size of $$ G_i\backslash F/C \cong q_i(C)\backslash q_i(F)/q_i(C) $$ which tends to infinity by construction.
On the other hand, since $g\in G_i$ for all $i$, clearly $G_igK=G_iK$.
