How does hashing twice protect against birthday attacks? The bitcoin wiki says:


  Bitcoin is using two hash iterations (denoted SHA256^2 ie "SHA256 function squared") and the reason for this relates to a partial attack on the smaller but related SHA1 hash. SHA1's resistance to birthday attacks has been partially broken as of 2005 in O(2^64) vs the design O(2^80). While hashcash relies on pre-image resistance and so is not vulnerable to birthday attacks, a generic method of hardening SHA1 against the birthday collision attack is to iterate it twice. A comparable attack on SHA256 does not exist so far, however as the design of SHA256 is similar to SHA1 it is probably defensive for applications to use double SHA256. And this is what bitcoin does, it is not necessary given hashcash reliance on preimage security, but it is a defensive step against future cryptanalytic developments. The attack on SHA1 and in principle other hashes of similar design like SHA256, was also the motivation for the NIST SHA3 design competition which is still ongoing.


I don't understand how hashing twice would prevent a birthday attack considering that finding a collision after the first hash would trivially entail a collision after the second hash. 

Is it because you have to compute the hash twice thus slowing the process down or is there a more significant reason?
A collision in any hash function gives a collision in a "squared" variant of the hash function. This is easy to see. If hash(x)==hash(y), then hashing the outputs will also be the same.

So the wiki entry is wrong. To see the real purpose of the double hash see this question and answers.