How does a range proof bound lower at 0 and not -1?
What if C and C' are commitments to 0 and -1? Then if I can sign, that means the ring signature is a proof is over {0,-1}.


To create a signature, the public key must be a known multiple of G. That multiple is known as the private key. When a point has a nonzero H term in it, no signature with that point as public key can be created. In order to write rG+vH as a multiple k of G, k = r + v*(G/H). H is constructed in such a way that nobody knows the ratio between G and H. As a result, rG+vH can only be a known multiple of G when v = 0.

As a result, you can only sign for public keys that are points which are commitments to the value 0.

A ring signature means you're picking multiple points as public key, and signing for 1 of them, without revealing which one.

So if C is a commitment to 0, and thus C' a commitment to -1, you absolutely can create a ring signature with {C, C'} as public keys. But it's due to the presence of C there.


  I'm probably missing something, so please tell me, why can't the amount be -1 in this proof?


There is no 'value' in the proof. There is a choice of two values. The signer only proves he can sign for one of them, and verifiers cannot know which one he meant.