Local operations and classical communication (LOCC) on quantum system

I assume the question refers to how LOCC is used to distinguish the two states $$ (|00rangle+|11rangle)/sqrt{2}qquad(|01rangle+|10rangle)/sqrt{2} $$ when Alice and Bob each hold one qubit, and are separated by a great distance.

There are many different measurement operators that could achieve the same task. If Alice just held both qubits herself, she could simply implement a measurement in the Bell basis, described by measurement operators $$ M_0=frac12(|00rangle+|11rangle)(langle00|+langle11|)qquad M_1=frac12(|00rangle-|11rangle)(langle00|-langle11|)qquad M_2=frac12(|01rangle+|10rangle)(langle01|+langle10|)qquad M_3=frac12(|01rangle-|10rangle)(langle01|-langle10|) $$ However, these measurements don't have a tensor product structure, and so cannot be implemented by LOCC. The results $M_0$ and $M_2$ distinguish the two states given.

As an alternative, Alice and Bob do as described at the original link; both make $Z$ basis measurements. So, their measurement operators are described by $$ M_0'=|0ranglelangle0|otimes |0ranglelangle0|qquad M_1'=|0ranglelangle0|otimes |1ranglelangle1|qquad M_2'=|1ranglelangle1|otimes |0ranglelangle0|qquad M_3'=|1ranglelangle1|otimes |1ranglelangle1| $$ If they get either answers $M_0'$ or $M_3'$, they had the first state, while if they get $M_1'$ or $M_2'$ they had the second state. Note that while Alice and Bob each act locally, on their qubit (as you can tell from the tensor product structure), they only know which overall result they got by comparing their measurement results, which requires classical communication.

Now a brief comment about the dimension of the Hilbert space. I have specifically talked about two qubits, $mathbb{C}^2otimesmathbb{C}^2$. The first set of measurement operators, ${M_k}$, cannot be described under this tensor product structure; they are operators on $mathbb{C}^4$, while the ${M_k'}$ operators can be described under $mathbb{C}^2otimesmathbb{C}^2$. As for why the original source was talking about $mathbb{C}^2otimesmathbb{C}^n$, for that specific example it seemed to be an unnecessary complication. Yes, you can always restrict to a qubit inside a larger Hilbert space, but there's no reason not to just take $n=2$ as far as I can see.