How does value of $p$ comes to $2^{512}$ for security complexity of
$2^{72}$?
Personally, I couldn't connfirm this number, but rather landed at $2^{66}$. If you don't know how to find this number, just plug the $2^{512}$ into the complexity equation given and take the logarithm to the base two for better readability.
Note that Schnorr's complexity estimate for the field is based on the Index-Calculus Algorithm and doesn't yet take into account the Number Field Sieve (original paper as PDF) against discrete logarithms, which would give a different expression, namely $L_p[1/3,3^{2/3}]=e^{(3^{2/3}+o(1))(\ln p)^{1/3}(\ln\ln p)^{2/3}}$, that would yield a much lower required attack complexity.
What is the equivalent AES and RSA key-lengths if we use $p$ of $1024$
bits and $q$ of $512$ bits?
$q$ is the subgroup order and if you get a 512-bit subgroup you'll have roughly 256-bit security for this subgroup. $p$ is the field prime and with it being 1024-bit, you'll get roughly 80-bit security against NFS-based attacks. So the answer is: A 1024-bit RSA modulus will have a comparable security along with AES-128, where 48 key bits are constant zero'ed out. Note that just plugging-in $2^{1024}$ into the complexity equations is a dangerous business, as $o(1)$ hides an implementation-dependenant (constant) factor that was not accounted for. Rather you need a concrete data point as a basis for further extrapolation. Here's a more detailed treatment of keylengths: (simple) (complex) by Lenstra and Verheul.
Since Schnorr is finite field signature scheme, what is the 'field
size', 'group size' and 'subgroup size' in case of schnorr
signature scheme?
$p$ is the field prime and its bitlength is the field size.
$q$ is the subgroup order and its bitlength is the subgroup size.
There's no relevant "group size" in this case (maybe one could say that's equivalent to the field size though)