That's flat out wrong. Working modulo 4 is not working in a finite field, because 22 = 0 when you work mod 4. When you work modulo a prime, then you are working in a finite field. Working in the finite field of 4 elements is not* the same as working mod 4. Finite fields of size p^n are not the same as (not isomorphic to!) the ring Z/(p^nZ) whenever n > 1.

Just because there exists a finite field of size 2^n doesn't mean that any length-n-bit-string that is the output of a hash function lives in a finite field of size 2^n.

Adding and multiplying hashes doesn't make any sense. If it does then you have a very weird hash function. Not one of the standard cryptographic hashes. So saying that a hash function maps onto a finite field sounds pretty confused to me.