M_{21} always = M_{12}. That can be shown by an energy argument.
It's correct I think to say that
M_{21} = N_{2}phi_{1}/i_{1}. That computes to
M_{21} = N_{2}A_{2}μn_{1}
where N_{2} is the number of turns in the inside coil 2. This agrees with your expression of M_{12} (I think you reverse the conventional sequence of the subscripts, but OK).
This is because the flux inside coil 2 is clearly very nearly the same as that inside coil 1.
It's more difficult to argue in the same manner to derive M_{12} because the question is how much of the flux generated by coil 2 really couples into coil 1.
The answer however has to be such that M_{12} = M_{21}. I have looked at several derivations of mutual inductance of various coil configurations and in each case they pick the easier computation for M_{ij} and then assume (correctly) that M_{ji} = M_{ij}.