I would do this carefully, the long way, and then see whether I have missed any shortcuts. A procedure that does not involve crossing the fingers has high priority!
You counted correctly the number $\binom{6}{3}\binom{4}{2}$ of teams made up of specialists. So there are $120$ such teams. Now we count separately the teams that contain $1$ versatile player, and $2$ versatile players.
If we are going to have $1$ versatile player, she can be chosen in $\binom{2}{1}$ ways. She can replace a forward, leaving $\binom{6}{2}\binom{4}{2}$ choices for the rest of the team, or she can replace a guard, leaving $\binom{6}{3}\binom{4}{1}$ choices for the rest of the team. Thus there are
$$\binom{2}{1}\left(\binom{6}{2}\binom{4}{2}+\binom{6}{3}\binom{4}{1}\right)$$
teams with exactly $1$ versatile player. So there are $340$ such teams.
If we are going to use $2$ versatile players, they can be both replace forwards, leaving $\binom{6}{1}\binom{4}{2}$ choices, or both replace guards, leaving $\binom{6}{3}$ choices, or do one of each, leaving $\binom{6}{2}\binom{4}{1}$ choices, for a total of
$$\binom{6}{1}\binom{4}{2}+\binom{6}{3}+\binom{6}{2}\binom{4}{1}$$
teams with exactly $2$ versatile players. So there are $96$ such teams.
Finally, add up.
Remark: I interpreted team to mean a set of $5$ people. If by team we mean a set of $5$, together with a specification of what positions (forward or guard) they are playing, the answer would be different, since one versatile player in each position would have to be counted twice, once for X playing forward and Y playing guard, and once for the reverse. And one can complicate things further.
Best Answer
Yes, this is correct. Another approach is to first choose the five players and then assign them to positions: $$\binom{12}{5}\binom{5}{2}\binom{3}{2}\binom{1}{1}=23760$$