My misunderstanding is probably due to a trivial thing I am not seeing.
In lattice surgery for surface code, we can perform a merging operation. This operation should be symmetric as indicated on Eq.5 in this paper, or Eq on top of page 5 of this one (and it is a natural consequences of the sequences of operation you perform on the surface).
More precisely, for a rough merge, calling the two states to be merged:
$$|\psi\rangle= \alpha |0\rangle + \beta |1 \rangle $$ $$|\phi\rangle= \alpha' |0\rangle + \beta' |1 \rangle $$
The merging operation acts on the logical space as:
$$|\psi\rangle \otimes |\phi\rangle \to \alpha |\phi \rangle + (-1)^M \beta X |\phi\rangle$$
We see that within the process, a two qubit state is mapped to a single qubit state: the merging operation does not preserve the dimensions. As far as I understand, the "net" process is somewhat equivalent to measure the $X_1 X_2$ observable of the two qubit state and to map the result to a single qubit state. The quantity $M$ is the measurement outcome of the observable $X_1 X_2$ on the initial two qubit state.
By construction, the merging operation is symmetric, hence, by exchanging the role of $|\phi\rangle$ and $|\psi\rangle$, we expect to have (as indicated in the papers provided):
$$ \alpha |\phi \rangle + (-1)^M \beta X |\phi\rangle=\alpha' |\psi \rangle + (-1)^M \beta' X |\psi\rangle $$
However, if you do the calculation you realize that this last equality is not true. Below is a short script illustrating the issue
How to make sense of this? On one hand I agree with the paper that we should have the equality true (because we are merging two different surface into a single one: the process should be the same if we exchanged the initial two surfaces). But on the other hand the concrete calculation indicates that it is not the case (and is then in contradiction with the papers).