From the blog post:

Most, but not all, people who have written about this are thirders. But:
On Sunday evening, just before SB falls asleep, she must believe the chance of heads is one-half: that’s what it means to be a fair coin.
Whenever SB awakens, she has learned absolutely nothing she did not know Sunday night. What rational argument can she give, then, for stating that her belief in heads is now one-third and not one-half?
As a stark raving Bayesian, I find this mildly disturbing.

The confusion is in dropping context. The context is of sampling the results of a coin flip, and in particular biased sampling based on the result of the coin flip.

Watching the flip on Sunday is a different sampling process than the Sleeping Beauty sampling process. That a biased sampling of a coin flip produces a bias in the observed outcomes should not be a shock to people.

Lets parameterize the Sleeping Beauty process by the number of awakenings in either path. SB(Hnum,Tnum). The standard process is SB(1,2). Let's consider the process SB(0,1) - on a flip of Heads, Sleeping Beauty is shot in the head and never wakes up, and on a flip of tails, Sleeping Beauty is woken up once.

P(Tails| awakening, SB(0,1)) = 1. Yes? Anyone not see that? This is not a cheat coin, this is Sleeping Beauty knowing she'll only awaken on a flip of Tails. Biased sampling. Just not complicated. Different process than the coin flip itself.

Similarly, P(Tails| awakening, SB(Hnum,Tnum)) = Tnum/(Hnum+Tnum).. More biased sampling.

Note that the blogger did not include an identification of the sampling process in the prior information he conditioned on in his equations, leaving him free to confuse the two different sampling processes he was thinking of.

But, just plug and chug the Jaynes way, conditioning on all your prior information, and voila! The result is transparent.

Jaynes wins again!