The counts do not seem to be right. If there is only $1$ person in the group, there are $0$ kisses.
If there are $2$ people in the room, there are $4$ kisses (A kisses B on both cheeks, $2$ kisses, B reciprocates, $2$ more).
Now look at the case of $3$ people, A, B, and C. Again we can make an explicit count. For A and B, we have $4$ kisses. For A and C, $4$ more, and finally for B and C, $4$ more, for a total of $12$.
If you have a certain amount of patience, you can deal with $4$ people, A, B, C, and D. First A exchanges kisses with B, C, and D. That's $12$ kisses. Now B, C, and D exchange kisses with each other. That's the case of $3$ people, which we have already solved! So they account for $12$ more, for a total of $24$.
Now deal with $5$ people, A, B, C, D, and E. First A does the kiss thing with B, C, D, and E. That's $16$ kisses. Then B, C, D, and E exchange kisses. That's already been computed, it is the $4$ people case, we get $24$. so the total is $16+24$. which is $40$.
Finally, deal with $6$ people, A to F. First A exchanges kisses with the $5$ others, total $20$. Then the remaining $5$ exchange kisses, total $40$. The sum is $60$.
To get a general formula, there are many approaches. We could follow up on our analysis above. Or we could start again.
For every pair of friends, there will be $4$ kisses. Do you know a simple formula for the number of pairs there are if there are $f$ friends in the group?
If you do not, here is a way of counting. We will call the people by the even more boring names $1,2,3,4,\dots,f$.
Let us first count the number of ordered pairs $(x,y)$ that we can form. The first person in the ordered pair can be chosen in $f$ ways. For each of these ways, the second person can be chosen in $f-1$ ways. So the total number of ordered pairs is $f(f-1))$. But this is not the number of plain pairs, for it counts (Alicia, Beti) as different from (Beti, Alicia). The fix is simple: divide our count of ordered pairs by $2$. So the number of pairs of people is
Now make your formula. To check for any mistakes, compare what your formula predicts with our explicit counts above.