It is used in to define the expected utility in the statement of these two theorems, eq. 27 and 30.

Yes. The point of those theorems is to prove that if your preferences are 'nice', then you are maximising Born-expected utility. This is why Born-expected utility appears in the statement of the theorems. They do not assume that a rational agent maximises Born-expected utility, they prove it.

The issue is that the agent needs a decision rule that, given a quantum state computes an action, and this decision rule must be consistent with the agent's preference ordering over observable macrostates (which has to obey the constraints specified in the paper).

Yes. My point is that maximising Born-expected utility is the only way to do this. This is what the paper shows. The power of this theorem is that other decision algoritms don't obey the constraints specified in the paper.

If the decision rule has to have the form of expected utility maximization, then we have two functions which are multiplied together, which gives us some wiggle room between them.

No: the functions are of two different arguments. Utility (at least in this paper) is a function of what reward you get, whereas the probability will be a function of the amplitude of the branch. You can represent the strategy of maximising Born-expected utility as the strategy of maximising some other function with respect to some other set of probabilities, but that other function will not be a function of the rewards.

Even if it turns out that it is, the result would be interesting but not particularly impressive, since macrostates are defined in terms projections, which naturally induces a L2 weighting. But defining macrostates this way makes sense precisely because there is the Born rule.

A macrostate here is defined in terms of a subspace of the whole Hilbert space, which of course involves an associated projection operator. That being said, I can't think of a reason why this doesn't make sense if you don't assume the Born rule. Could you elaborate on this?