You are objectively wrong. The math is straightforward to show that you can operate on a ciphertext securely in some cryptosystems.

Consider two integers M1 and M2.

Consider RSA with private key (E), public key (D), and public modulus (N).

Encrypt(M, E, N) = mod(pow(M, E), N).

Decrypt(C, D, N) = mod(pow(C, D), N).

mod(Encrypt(M1, E, N) * Encrypt(M2, E, N), N) = mod(Encrypt(M1 * M2, E, N), N).

So, for all RSA encryption, multiplying the ciphertexts results in a ciphertext that is the multiple of the plaintexts. However, unless you can break RSA, you can not determine what numbers you multiplied or what the final multiplied number is.

This is not a fully homomorphic system as it only allows multiplication, but it is a existence proof that you can do operations on ciphertext that apply to the plaintext without being able to recover the plaintext unless you can break the encryption directly.

This is actually the magic of FHE. The ciphertext is indistinguishable from noise AND can be computed on, it just looks like different noise.

If you believe the underlying cryptographic hardness assumption of LWE/RLWE/etc, then yes Google cannot see any of the input or output of the model.

they could have gone with an oblivious transfer approach (where it's working on what looks like multiple problems at once, you don't know which)

Eh? With secret sharing one can do computation on a shared secret where it is provable that no individual party can recover any information about the data with their share alone.

I don’t see why you conclude that FHE couldn’t be close to as secure as that. (Like, not information theoretically, but with computationally bounded adversaries.)

No, that’s not what proper encryption means. Security for encryption means that cipher texts encrypting distinct messages are indistinguishable. This is called IND-CPA, and FHE satisfies this.