Wouldn’t P(you wake up | heads) be 1⁄2, since you wake up on Monday but not on Tuesday? That would give P(you wake up) = 3⁄4, which gives P(heads | you wake up) = 1⁄3. That is, “you wake up” is not taken to mean “you wake up at least once,” but “you wake up today (where today is Monday or Tuesday).” P(heads | you woke up at least once during the two days) is obviously 1⁄2.
I think not waking up/getting asked on HeadsTuesday (i.e. the Heads was flipped and the date is Tuesday) is important. This is the information you “gain” when you wake up (oh hey, it’s not HeadsTuesday). When you are answering on Sunday, you answer for all outcomes, including when you don’t wake up, but “given you’re awake” excludes HeadsTuesday.
Let A, B, C, D denote HeadsMonday, HeadsTuesday, TailsMonday, TailsTuesday, and W = {A, C, D} be the set of days you wake up. We need to assign values P(A), P(B), P(C), P(D) with sum 1, and are looking for P(H | W) = P(A) / (P(A) + P(C) + P(D)). From the coin being fair, P(H) = P(A) + P(B) = P(T) = P(C) + P(D) = 1⁄2.
If you think the chosen date and the coin flip are independent then you must have P(A) = P(B) = P(C) = P(D) = 1⁄4, so P(H | W) = 1⁄3. If not, the only way to get P(H | W) = 1⁄2 is for P(A) = 1⁄2, P(B) = 0. So, if Tuesday simply doesn’t exist after flipping heads, you could be able to get P(H | wake up) = 1⁄2.
Given the above formulation, the inference is not the same. P(H) = 1⁄2, but after information about W or not W is given, then P(H | W) = 1⁄3 or P(H | not W) = 1. The math doesn’t care, you just aren’t awake to perform your update process. When precommitting, you are not manipulating P(H), you are manipulating P(H |W) by changing W, so there’s no issue.
P(A) = 1⁄2, P(B) = 0 is still the only way I can see to get P(H | W) = 1⁄2. In which case, I can’t find any non-artificial framing for why Heads Tuesday does not exist (and Heads Monday exists twice as much as Tails Monday).