Assuming my simple example is correct, I think the answer to the 4 couples problem is that your wife shook hands with 4 people.
Here is my reasoning:
In the simple example with one couple, A1 and A2, A1 could have shaken hands with W and Y, and A2 with no one, or vice versa. In either case, W shook hands wth one person.
Extend this example to 2 couples, A1, A2, B1, B2, plus W and Y.
One solution is for A1 to shake hands with B1, B2, W and Y (4 people), A2 to shake hands with no one, B1 to shake hands with W and Y (2 people, plus A1), B2 to shake hands with no one else (plus A1).
So A1 shakes hands with 4 people, A2 with 0, B1 with 3, B2 with 1, and W with 2. I think A1, A2, B1, B2 could be various permutations, but W will always be 2.
Extend this to 3 couples and one solution is
A1 with 6, A2 with 0, B1 with 5, B2 with 1, C1 with 4, C2 with 2 and W with 3. Again, various permutations of A1, A2, B1, B2, C1, C3 but W always with 3.
Extend this to 4 couples and W will shake hands with 4 people.