This is really a comment, but it's too long:
I'm not entirely sure what you're asking, but I think there are a couple things worth pointing out:
In terms of what a model $M$ "knows" about a structure Ackermann-coded into one of its subsets $A$, note that this question best makes sense if $M$ knows about $A$ to begin with. But most of the time, the $A$s we'll be interested in won't be definable in $M$: indeed, no model $W$ of ZF can be Ackermann-coded into any nonstandard model $M$ of PA in a definable way (since such an encoding would give evidence in $M$ of the non-standardness of whatever element of $M$ is coding the $\omega$ of $W$). So you have to be very careful when you ask "What does $M$ know?" about such a situation - what exactly do you mean, given that the most basic information about such a configuration will generally be undefinable in $M$?
In terms of Parikh's model, I think you're overestimating its relevance to Ackermann coding. Any model of PA has a unique definable map satisfying the recursive properties of exponentiation. (Indeed, PA proves "there is a unique exponentiation function" in the following sense: PA proves "for all $n$, there is a unique map from $\{0, 1, . . . , n\}^2$ to $\mathbb{N}$ which satisfies [recursive properties]". Note that such a map is a finite object, so can be talked about directly in PA, even though the whole exponentiation map can't.) Ackermann coding works with respect to this definable map. Parikh showed that we can have undefinable maps that also satisfy the properties of exponentiation. But these maps aren't relevant to Ackermann coding.
Both bullet points really fall under the same category: although "strong foundational" theories (like PA and ZF) imply lots of rigidity about the universe in various ways, their nonstandard models have tons of flexibility, undefinably. For example, every nonstandard model $W$ of ZF contains a set model $m$ of ZF, even if it proves that ZF is inconsistent! . . . the trick being, of course, that the model is incorrect about what ZF is, and $m$ fails to satisfy some of the axioms of ZF$^W$, so the sense in which $m$ is a model of ZF is really undefinable. Similarly, Cohen's result that we can have a model of ZF with an automorphism of order $2$ is inherently about the undefinable structure of models, since no model of ZF has any definable automorphisms at all.
Ressayre's substructures and Parikh's exponentiations are examples of this "undefinable flexibility". But because they're appropriately undefinable, they're quite slippery, and don't interact with definable phenomena (like Ackermann coding) in any nice way.