In the first problem there are $\binom{12}{2,2,2,2,2,2}$ ways for the teachers to choose $2$ children each without restriction. There are $6\binom{10}{2,2,2,2,2}$ ways for the first teacher to choose a pair of twins and the other five teachers to choose without restriction, and the same goes for each of the other $5$ teachers, so the first correction in the inclusion-exclusion count is
$$\binom{12}{2,2,2,2,2,2}-\binom61\cdot6\binom{10}{2,2,2,2,2}\;.$$
There are $6\cdot5\binom8{2,2,2,2}$ ways for any given pair of teachers to choose a pair of twins each, and there are $\binom62$ pairs of teachers, so the second correction term results in
$$\binom{12}{2,2,2,2,2,2}-\binom61\cdot6\binom{10}{2,2,2,2,2}+\binom62\cdot6\cdot5\binom8{2,2,2,2}\;.$$
I expect that you can probably finish the inclusion-exclusion argument from here.
For this particular problem there is an easy way to count (but right now I don't see how to generalise this to more than $4$ couples). One can first seat the women in $4$ alternating seats. Assuming you mean to identify rotationally symmetric arrangements this can be done in $3!=6$ ways: the first women serves as reference and her seat can be numbered 0, and seating the three other women is given by a bijection to the seats $2,4,6$. (If you also want to identify reflection symmetry, divide by $2$.)
Now to seat the men, there are two options for the husband of the lady in seat $0$, namely seats $3$ and $5$. But when this is done, the arrangement is fixed. Supposing he took seat $3$, then this seat is no longer available for the husband of the lady in seat $6$, who then must go to seat $1$; then the husband of the lady in seat $4$ must go to seat $7$, and the remaining husband to seat $5$. In case the first husband took seat $5$, the situation is similarly fixed, reasoning in the opposite direction. So in all there are $6\times 2=12$ solutions.
Added I finally found out that for $n$ couples the number is given (up to a factor $(n-1)!$ for seating the women first) by A000197 in OEIS. Presumably you can find useful things in the comments and formulas there.
Best Answer
THIS IS WRONG (SEE COMMENTS). I'M LEAVING IT UP AS AN EXAMPLE OF A POSSIBLE METHOD.
Let's call the people Mr. and Mrs. A, B, and C. Let's number the seats at the table clockwise 1 through 6, and say that Mr. A always sits in seat 1. (As the table is circular, seating patterns count as identical if they differ by rotation, so every seating pattern is equivalent to a seating pattern with Mr. A in seat 1.) We know Mrs. A can't be in seat 2 or 6. Now let's break into cases, depending on whether the occupants of seats 2 and 6 are married.
Case 1: the occupants of seats 2 and 6 are married. Four symmetrical sub-cases: Mr. B is in seat 2 and Mrs. B is in seat 6 (or vice versa), or Mr. C is in seat 2 and Mrs. C is in seat 6 (or vice versa). Let's take the first sub-case for concreteness. Then Mrs. A and Mr. and Mrs. C have to occupy seats 3, 4, and 5. Mrs. A has to take seat 4 (otherwise Mr. and Mrs. C would sit next to each other), so there are two ways to arrange seats 3, 4, and 5 (either Mr. C takes seat 3, or Mrs. C does). The other three sub-cases are identical, giving $2 \times 4 = 8$ seating patterns.
Case 2: the occupants of seats 2 and 6 are unmarried. Four symmetrical sub-cases: Mr. B in seat 2 and Mrs. C in seat 6 (or vice versa), or Mrs. B in seat 2 and Mr. C in seat 6 (or vice versa). Let's take the first case for concreteness. Mrs. A, Mrs. B, and Mr. C are in seats 3, 4, and 5, with the restrictions that Mrs. B can't take seat 3 and Mr. C can't take seat 5. In this case, any of the three can take seat 4, but the occupants of seats 3 and 5 are determined by the occupant of seat 4 (because Mrs. B and Mr. C must take seats 5 and 3, respectively, if they are not chosen for seat 4). This gives three patterns per sub-case, so $3 \times 4 = 12$ overall.
The total number of possibilities is $8 + 12 = 20$.