Why are there $q^2$ keys for each equivalence class in Pedersen-van Heyst Signature Scheme? How do you know there are $q^2$ keys for each equivalence class in the Pedersen-van Heyst Signature Scheme? My textbook says this is "easy to see", however, I am not seeing it.

Is it because in the signature  $Sig_K(\gamma_1,\gamma_2)$, the parameters $\gamma_1$ and $\gamma_2$ are both elements in $\mathbb{Z}_q$?
I did not know the scheme, so I had to google it to find the missing part of its description - here it is, in case someone else wants it.

So, two keys are in the same equivalence class when they have the same $\gamma_1$ and $\gamma_2$, id est, when $\alpha^{a_1}\beta^{a_2} = \alpha^{a'_1}\beta^{a'_2} \bmod p$, and $\alpha^{b_1}\beta^{b_2} = \alpha^{b'_1}\beta^{b'_2} \bmod p$, with $p = 2q+1$. Note that the exponents of $\alpha$ "live" in a cyclic group of order $q$ (as I do not know how familiar you are with cyclic groups, I assume this statement is clear for the moment, don't hesitate to ask if it is not the case). If we denote by $s$ the discrete logarithm of $\beta$ in base $\alpha$ modulo $p$, we can rewrite the "same $\gamma_1$ and $\gamma_2$" condition as follows:

$a_1+ sa_2 = a'_1 + sa'_2 \bmod q$ and $b_1+ sb_2 = b'_1 + sb'_2 \bmod q$

Now it should become more clear to you: we have exactly two equations modulo $q$, and four variables ($s$ is uniquely determined by $\beta$ given $\alpha$), hence exactly $q^2$ solutions. Does that make sense to you ?

EDIT: regarding your last question, $\gamma_1$ and $\gamma_2$ are group elements, they are in the group of order $p$. Only the exponents (which correspond to the secret key, $\gamma_1$ and $\gamma_2$ being the public key) are elements of $\mathbb{Z}_q$.