Is the universal wavefunction globally coherent?

Considering only the many-worlds interpretation of quantum theory.

You can think of the universal wavefunction as a pure state (and if it is somehow not, just add qubits until it is one) and always stays that way. So if you have a wavefunction of the form $$|Psirangle = frac{1}{sqrt{2}} left(|phi_{1}rangle + |phi_{2}rangle right)$$ then you can find $|phi_{1}rangle$ and $|phi_{2}rangle$ can interfere with each other just like normal.

When you start to think about observers it gets a bit more confusing but writing the universal wavefunction as: $$|Psi(t)rangle = frac{1}{sqrt{2}} left(|o_{1}(t)rangleotimes |s_{1}(t)rangle + |o_{2}(t)rangleotimes |s_{2}(t)rangle right).$$ Then the question becomes, can the systems $s_{j}$ interfere with each other, and the answer is yes, but only if/when the two observers match up with each other $$|o_{1}(t^*)rangle = |o_{2}(t^*)rangle.$$

If this happened then regardless of which path you took you'd have exactly the same thoughts at this time. It would also appear that this should only ever happen instantaneously, however when we are at times near $t^*$ we can always express $|o_{j}rangle$ as some sum of the observer's state at the critical time $|0rangle$ plus some small perturbation by state $|jrangle$ that goes to zero as $trightarrow t^*$.

This argument is fairly simplified as the observer is made of well beyond trillions of qubits and so you probably don't have to worry about this looping procedure occurring and instead will only ever see interference if you can keep the coupling between the observer and the system sufficiently small (and so don't see interference arising due to interfering branches).