What happens when we hash already hashed values, concatenated together?

I read on the page 16 of On the Security of Hash Function Combiners that

the classical combiner for collision-resistance simply concatenates the outputs of both hash functions $Comb_{\mathbin\|}(M) = H_0(M) \mathbin\| H_1(M)$ in order to ensure collision resistance as long as either of H0, H1 obeys the property.

Consider H, a secure internal hash function with 256-bit inputs and 128-bit outputs

My question is, if we hash the concatenation again with the same hash function used before,

Like so

H(H(M)∥H(M))) (pardon my lack of knowledge of what latex is)

what happens to the collision resistance? does it become

$$2^{(n/2)} \cdot 2^n$$

or does it become

$$2^{(n/2)} + 2^n$$

or am I going the wrong direction to find if it even improves or makes it worse? The hash function used all three times is the same hash function, which I am assuming to be ideal/robust and collision resistant.