Presumably everyone is to be seated in a line. The $k$ tutors divide the students into $k+1$ possibly empty groups. Each of the internal groups must contain at least one student so that the two tutors on the boundary of the group are not seated next to each other. This accounts for $k-1$ of the students, so we need only divide the remaining $n-k+1$ students into $k+1$ groups. This can be done in
$$ \binom{n+1}{k} $$
ways.
(Imagine seating the remaining students in a line and placing the tutors between them. There are $n-k+1 + k = n+1$ people in total, and we must choose which $k$ of them are tutors.)
Edit:
The reason why your answer is incorrect is that in the case where the first person is a tutor, it is indeed the case that the second person must be a student. But this then accounts for one tutor and one student, so there are now $n-1$ students and $k-1$ tutors which must be placed, which can be done in $F(n-1, k-1)$ ways, not $F(n-1, k)$ ways. You forgot to decrease the number of tutors which must still be placed.
Similarly, if the first person is a student, then there are no restrictions. There are now $n-1$ students and $k$ tutors, and the only restriction is that two tutors can not be seated next to each other, so there are $F(n-1, k)$ ways in which the people can be seated in this case. This gives us that
$$ F(n, k) = F(n-1, k-1) + F(n-1, k). $$
If you want to further divide up the case in which the first person is a student into the cases where the second person is a student, and where the second person is a tutor, then you must remember that if the second person in a tutor, then the third person must be a student, so there are $F(n-2, k) + F(n-2, k-1)$ ways in which the people can be placed in this case, not $F(n-1, k) + F(n, k-1).$
