The problem can be interpreted in more than one way. For the sake of simplicity, we think of one of each couple as female, and of the other as male.
We interpret the problem as follows: How many ways are there to pick $4$ people, $2$ of each sex, so that no two people picked are a "couple," and then divide these $4$ people into two gender-mixed teams of $2$?
The females can be picked in $\binom{7}{2}$ ways. For each of these ways, we can pick the males in $\binom{5}{2}$ ways, for a total of $\binom{7}{2}\binom{5}{2}$. Once we have done the picking of the $4$ people, they can be divided into two mixed teams of two in $2$ ways, giving a total of $2\binom{7}{2}\binom{5}{2}.$ This turns out to be $420$. To make the answer of $840$ correct, we would need to find another interpretation of the problem. The interpretation that having a couple in the foursome is OK, as long as they are not on the same side, gives a number well below $840$. We do get $840$ if the only thing we forbid is two couples, but that would be a very strange interpretation of the wording of the question.
Edit: In response to the edited version, line up the "females" in order of height, or social Insurance number. Then line up the males opposite them. Producing the $7$ teams is equivalent to producing the derangements of $7$ objects. This number can be computed in various ways. It is substantially larger than $840$.