Does the double-hash H(H(x)) have greater collision probability than H(x)?
I'm not sure what the question here is, but obviously applying the hash function twice can never decrease the number/probability of collision as all collisions in the first invocation are maintained. 

However if H is collision free( a permutation as opposed to a random function) doubling will not cause any more collision it will remain collision free. So we see the number of collision does not strictly increase. 

For a PRF, the number of collision does clearly increase, this is one reason why when we iteratively hash passwords in for key derivation we mix in the input again every round and not simply hash calculate $H^n(salt || pass)$ 

If you look at the structure of a function you will see several cycles and threads(trees) leading into these cycles:


The cycles do not generate more collisions when doubled. Each thread meeting a cycle contributes one collision, If All the threads are of length 1 no more collisions are added when applying twice in any other case the number of collisions increases.