What?
Just as a string of individual digits makes a larger number.
However, that doesn't really address the original question on how much harder cracking 2x64bit hash would be than cracking a single 128bit hash would be.
My best guess there would be that it's really quantify as you start opening up more dimensions besides the number of bits. The gain would mostly come from protection against other properties from one of the algorithms like a potentially hidden backdoor or a undiscovered mathematical weakness. So as long as the strength of the individual hash function holds up, it probably makes sense to diversify between hash functions. E.g. SHA3-256 + BLAKE3-256, probably offers better long-term security properties than using SHA3-512.
This is not true. Consider two hash functions f and g
f(x) = md5(x)[0..63]
g(x) = md5(x)[64..127]
h(x) = f(x) || g(x)
So no, concating multiple smaller hash functions is not any weaker than using a single big one.
I have to admit that I'm not the greatest cryptography whiz, but I can't image that this holds up for _independent_ hash functions, where you should be able to more cheaply run a preimage attack against two 64bit hash functions than one 128bit hash function.
performance is probably an issue. SHA256 has 256 bits and SHA1 has 160 bits (1.6x more bits), but SHA256 isn't 1.6x slower, it's only 38% slower. benchmarks used: https://www.cryptopp.com/benchmarks.html
Back to the original question of "how secure are 2x 64bit hashes compared to 1x 128 bit hash?", I can't imagine how it could be any more secure, considering that if it were more secure, you could just make your 128 bit hash function be the concatenation of the two 64 bit hash functions. It might be equally secure, but I'm not sure why you'd use it over a properly designed 128 bit hash function.
Try doing that in my example.
I appreciate you making me think deeper about hash functions. Nice construction.