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.
Label the couples A, B, C so that the six people are A$_{1}$, A$_{2}$, B$_{1}$, B$_{2}$, C$_{1}$ and $C_{2}$.
Now, consider your 6 seats: _ _ _ _ _ _
There are 6 choices for who can sit in the first seat; without loss of generality, say it's A$_{1}$.
There are 4 choices for who can sit in the second seat (anyone else but A$_{2}$); without loss of generality, let's say this is B$_{1}$.
Our table now looks like this: A$_{1}$ B$_{1}$ _ _ _ _
For the 3rd seat, we just can't have B$_{2}$, so we can either have A$_{2}$ or one of the C's; I'll split this into 2 cases.
Case 1: A$_{2}$ sits in the third seat.
A$_{1}$ B$_{1}$ A$_{2}$ _ _ _
Now the fourth seat can't be B$_{2}$, otherwise the C's will sit next to each other in the last two seats. So the fourth seat must be one of the C's; there are 2 choices for this. Then the fifth seat must be B$_{2}$ and the sixth must be the other C.
Thus there are $6*4*1*2*1*1 = 48$ ways for Case 1 to happen.
Case 2: One of the C's sits in the third seat; there are 2 possibilities for this. Let's say it's C$_{1}$
A$_{1}$ B$_{1}$ C$_{1}$ _ _ _
Then there are 2 choices for the fourth seat: A$_{2}$ or B$_{2}$. I'll treat these as subcases.
Case 2.1: A$_{2}$ sits fourth.
A$_{1}$ B$_{1}$ C$_{1}$ A$_{2}$ _ _
For the fifth seat, we have 2 choices, and then 1 for the sixth.
Thus $6*4*2*1*2*1 = 96$ possibilities for this case.
Case 2.2: B$_{2}$ sits fourth.
A$_{1}$ B$_{1}$ C$_{1}$ B$_{2}$ _ _
For the fifth seat, there is only one possibility: A$_{2}$, since otherwise A$_{2}$ would be in the sixth seat, which is adjacent to the first (the table is circular). Of course there is only one possibility for the sixth.
Thus there are $6*4*2*1*1*1 = 48$ ways for Case 2.2 to happen.
There are then $48+96+48 = 192$ possible seating arrangements. The probability of this happening is $\frac{192}{6!} = \frac{192}{720} = \frac{4}{15} \approx 0.267$
Best Answer
A broad hint, not a complete solution
I could not follow your answer, but it might be worth checking it with an example like $k = 3, n = 1$ to see whether it works.
Rather than placing the husband first, I'd say that you have $n$ items of one type (couples) $C$ and $2(k-n)$ items of another type, singles, $S$.
Label the seats of the table $1, 2, 3, \ldots, 2k$. A seating plan then consists of
For the first bullet, a stars-and-bars kind of solution is probably what you want to think about.