the tag depends only on the checksum (that is just the binary sum of plain blocs)
This is technically correct but overlooks a crucial point: the "plain blocs" are the plaintext obtained by decrypting the ciphertext. They are not plaintext supplied by the attacker.
OCB is an AE(AD) mode. The security model for these modes is that the attacker must not be able to submit valid ciphertext (without reference to any particular plaintext). This is not the same as the security model for MACs, which is that the attacker must not be able to submit an existential forgery, i.e. a valid tag for a plaintext chosen by the attacker.
So you are quite correct that an attacker can take a known plaintext/ciphertext pair, modify plaintext so that it XORs to the same value as the known plaintext, and it will result in the same tag when encrypted under OCB. However, the OCB ciphertext will obviously be different! And the attacker has no way to generate this different ciphertext corresponding to the modified plaintext.
This underscores an important difference between the AE(AD) modes and the Encryption+MAC paradigm: with AE(AD) the verifier must always process the entire ciphertext, even if they independently receive (what they think is) the valid plaintext (because it may have been chosen by a helpful attacker who wants to save you the trouble of doing the decryption).
Bottom line: an OCB tag is not a MAC!
(Indeed this is obvious from the specification of the AD part of OCB: AD is of course not encrypted, and therefore cannot be decrypted to feed into the tag calculation together with the plaintext. Instead AD is processed through the cipher and enters the tag calculation as ciphertext, not plaintext).
I think this is the correct answer to the question. The input into the tweak of the tag calculation, as proposed in the other answer, is a red herring.