Schnorr protocol: how does malicious verifier win?
This all seems to be explained in the book (warning: large) you mention, all I mention here is taken from there. 

This is the last sentence before $\S18.3.2$:


  Enlarging challenge size not only improves performance (a positive result), in $\S18.3.2$ we will further see that this also has a negative consequence. Be careful, size matters!


This last sentence is of critical importance, but we will focus on the first one for now. Moving to $\S18.3.2$, we find the last sentence before $\S18.3.2.1$:


  However, there is a subtlety for the problem. Let us examine it now.


This is followed by about 2 pages of explanation to your question, which indeed they agree to have subtleties. Although you have read all the internet twice (kudos), your answer suggests you may have missed this.

It seems that setting 


  $${\tt Challenge}={\tt prf}(M||{\tt Commit})={\tt prf}(M||g^{\tt Response}y^{\tt
&gt; Challenge}\bmod{p})$$


leads to two options:


  
  The equation was constructed by Alice using her private input, and hence Alice discloses the fact that she has been in interaction with,
  and fooled by, $\tilde{Bob}$, or
  $\tilde{Bob}$ has successfully broken the pseudo-random function ${\tt prf}$ of the large output space $\mathbb{Z}_q$, because he has
  constructed equation $${\tt Challenge}={\tt prf}(\cdots y^{\tt Challenge}).$$ This is a well-known hard problem because ${\tt prf}$
  is assumed to be one-way.
  
  
  Given that $\tilde{Bob}$ is assumed to be polynomially bounded, the
  third party will of course believe that (i) is the case.


The information revealed is thus the fact that the prover has been interacting with the malicious verifier. Luckily, the authors mention that


  A small consolidation for Alice is that the information disclosure caused by $\tilde{Bob}$ is not a too disastrous one (though this assertion has to be based on applications really). [...] the proof transcript still does not disclose to $\tilde{Bob}$ or a third party any information about Alice's private input $a$.