Works for this simple case, but I think iterating the procedure is more general. For instance, suppose there are 6 people -- A1, A2, A3, B1, B2, B3, and you want it to be recoverable given any 2 of the A group and any 2 of the B group; I'm pretty sure that can't be handled with a solution like the above.
I think your right. One (k,n)=(2,3) scheme for each group to unlock the (2,2)-scheme for the secret.
And the wife just needs to remember one key this way...
Simplest would just be to have the 3-of-50 decrypt to another encrypted file, and give the wife the passphrase to that. Nobody holds multiple keys, satisfies the 'AND', and is simple.