I think it is explained in your OP. We do the same thing with a bit more detail.
Since we have already counted the number of "bad" positions with all the boys together, it remains to count the number of bad positions in which the boys are not all together, but some boy is not next to a girl.
There must be two boys together, and they must be at the left end or the right end ($2$ choices). Take such a choice, say left end. Then the remaining boy can be in any one of $3$ places, namely positions $4$, $5$, or $6$. That gives us $(2)(3)$ ways of choosing the seats the boys will occupy. For each of these $(2)(3)$ choices, there are $3!$ ways of permuting the boys among the chosen seats, and for every such choice there are $3!$ ways of permuting the girls, for a total of $(2)(3)(3!)(3!)$.
Another way: Counting all possible arrangements and subtracting the "bad" ones is often good strategy. But let us count directly.
Our condition will be satisfied if either no two boys are together, or exactly two boys are together and they are not in the end seats.
Count first the arrangements in which no two boys are adjacent. Write down $G\quad G\quad G$. This determines $4$ "gaps" into which we can slip the boys, one boy per gap. There are $4$ "gaps" because we are including the end gaps. There are $\binom{4}{3}$ ways of choosing $3$ of these gaps.
Or else we could slip $2$ boys into one of the two center gaps ($2$ choices), and then slip the remaining boy into one of the $3$ remaining gaps, for a total of $6$ choices.
Thus the places for the boys can be chosen in $10$ ways. For each of these ways, we can arrange the boys is $3!$ ways, and then the girls in $3!$ ways, for a total of $360$.
You are implicitly assuming that there are 13 seats for $6+6=12$ people. It is likely that the original question is making the implicit assumption that there are only 12 seats. In this latter case, when a boy sits first, there $6! \cdot 6!$ ways to sit the others. Multiply this by 2 to cover the case where the first seater is a girl.
Best Answer
The number of arrangements in which exactly one boy sits between the girls is $$12 \cdot 2! \cdot 12!$$ since there are twelve ways to choose the boy who sits between the girls, two ways of choosing the girl who sits to his left, one way of choosing the girl who sits to his right, and $12!$ ways of arranging the block of three people and the other eleven boys.
The number of arrangements in which exactly two boys sit between the girls is $$12 \cdot 11 \cdot 2! \cdot 11!$$ since there are twelve ways to choose the boy who sits in the first seat between the two girls, eleven ways to choose the boy who sits in the second seat between the two girls, two ways to choose the girl who sits to their left, one way of choosing the girl who sits to their right, and $11!$ ways to arrange the block of four people and the other ten boys.
Notice that $$14! - 2!13! - 12 \cdot 2!12! - 12 \cdot 11 \cdot 2!11! = \binom{11}{2}2!12!$$ in agreement with the answers provided by drhab and Henning Makholm.