What sort of language does the language of second-order arithmetic become if the 'numbers' are the finite ordinals?

Question

The language of second-order arithmetic is defined as follows (the wording of this definition is due to Henry Towsner from a pdf file of "[Chapter 4], "Second Order Arithmetic and Reverse Mathematics" I found on the internet):

Definition 4.1. The language of second order arithmetic is a two-sorted language: there are two kinds of terms, numeric terms and set terms.

0 is a numeric term,

There are infinitely many numeric variables, $x_0$, $x_1$,..., each of which is a numeric term.

In $s$ is a numeric term, then $\mathbf S$$s$ is a numeric term,

If $s$, $t$ are numeric terms then $+$$s$$t$ and $\cdot$$s$$t$ are numeric terms (abbreviated $s$+$t$ and $s$$\cdot$$t$),

There are infinitely many set variables $X_0$, $X_1$,..., each of which is a set term,

If $t$ is a numeric term and $S$ then $\in$$t$$S$ is an atomic formula (abbreviated $t$$\in$$S$),Tf $s$,$t$ are numeric terms then $=$$s$$t$ and $\lt$$s$$t$ are atomic formulas (abbreviated $s$=$t$ and $s$$\lt$$t$).

The formulas are built from the atomic formulas in the usual way.

Since it is known that $ZF$-Infinity (that is, $ZF$ woth the axiom of infinity dropped--I will not add the alternate axiom $\lnot$Infinity since the fragment alone will do all I need) derives all of the theorems of $PA$, it might be interesting to see how to formulate second-order arithmetic in terms of that fragment.

Obviously, some adjustments in the language will have to be made since the 'numbers' will now be finite ordinals. For example, the phrase 'numeric term' would have to be replaced by 'set term' (as in "$\emptyset$ is a set term, 'If $s$ is a set term, then $\mathbf S$$s$ is a set term' (where $\mathbf S$$s$ will have to be defined as $s$$\cup${$s$}); '$+$', '$\cdot$' will have to be replaced by the set-theoretic analogues for finite sets and '$\lt$ will have to be replaced by $\subset$).

But now what to do with the phrase 'set term' in 'There are infinitely many set variables $X_0$, $X_1$,..., each of which is a set term', since for $ZF$ -infinity one already has '$\in$' (for example) defined for the finite sets, and Separation can be used to define properties of finite ordinals analogous to properties of natural numbers in $PA$?

Question 1: Can the axioms of $ZF$-Infinity be so adjusted that the language of second-order arithmetic need only one sort--the sort set?

There is, of course, a more natural way to adjust the language of second-order arithmetic so that it is still a two-sorted language:

Replace the term 'numeric term' with 'set term', and 'set term' with 'class term', where 'set' can now be defined as follows:

'$x$ is a set' $\leftrightarrow$ $\exists$$X$($x$$\in$$X$), where $X$ is a class variable.

Given this definition of set, it might be reasonable to allow sets to be classes that can be members of other classes and in this fashion reduce the number of sorts of the language of second-order arithmetic from two to one. From this also one can see the role of the Comprehension axioms in second-order arithmetic (the usualway of defining class--as extensions of formulas). As regards second-order induction ($\forall$$X$($\emptyset$$\in$$x$ $\land$ $\forall$$x$($x$$\in$$X$ $\rightarrow$ $x$$\cup${$x$}$\in$$X$) $\rightarrow$ $\forall$$x$$x$$\in$$X$)), it allows for the existence of infinite classes (which can be deemed sets if such classes can be members of other classes--this is the true significance of the axiom of infinity).

Question 2. Does the 'bi-sorted' language described above (with a 'set' sort and a 'class' sort, with a class being a set iff that class can be a member of some class) seem a better way of adjusting the language of second-order arithmetic if $PA$ is replaced by $ZF$-Infinity than the alternative in Question 1 ?

(Note: It might be that neither of the alternatives presented are suitable for the task; so in that case, how does one properly define the language of second-order arithmetic when $PA$ is replaced by the fragment $ZF$-Infinity? [Question 3]

Answer 1

Your question $2$ has the right idea. The right version of "second-order finite set theory" is two-sorted, as is second-order arithmetic, with one sort being the "class sort." (Note that in a sense, hereditarily finite sets will show up both as "numbers" and as "sets," but this is no different from the fact in second-order arithmetic that a finite set can be coded by a single number; it's just a bit more striking here because the finite set of hereditarily finite sets literally is a hereditarily finite set.)

Now an important point here is that second-order arithmetic is much stronger than PA! For a really easy example of this, it implies the consistency of PA. So not every model of finite set theory can be expanded to a model of second-order finite set theory, identically to how not every model of PA can be expanded to a model of second-order arithmetic. That said, by restricting the comprehension axioms, we can get a second-order finite set theory which is conservative over the first-order finite set theory (identically to how, by restricting comprehension in second-order arithmetic, we can get ACA which is conservative over PA).

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Note that this is exactly what we should expect: by the analogy "numbers $\sim$ hereditarily finite sets,", we expect "sets of numbers $\sim$ sets of hereditarily finite sets". So that's the language setup we should be looking for. Remember, the whole point of this analogy is that - once we're looking at sufficiently strong theories on each side - there is no real difference between finite set theory and arithmetic.