This is just classic stars and bars problem. The solution is:
$$\binom{n+m-1}{n} = \binom{8+4-1}{8}$$
if it's allowed a a school to have no teacher. While if a school must have at least one teacher the solution is:
$$\binom{k-1}{n-1} = \binom{7}{3}$$
For better explantion on this method you can read about it one the Wikipedia page. It has a really nice explanation.
I would do the problem exactly the way you did it.
Here is an alternative approach to confirm your answer.
We can line up the five students in $5!$ ways, leaving spaces between them and at the ends of the row in which to insert the teachers. There are six such spaces, four between successive students and two at the ends of the row. We can insert the three teachers in $P(6, 3) = 6 \cdot 5 \cdot 4$ ways. This gives us $$5! \cdot 6 \cdot 5 \cdot 4$$ linear arrangements of students and teachers in which no two teachers are consecutive.
However, since we wish to arrange the students and teachers around a circular table so that no two teachers sit in consecutive seats, we must exclude those linear arrangements in which teachers are at both ends of the row. There are three ways to select the teacher at the left end of the row, two ways to select the teacher at the right end of the row, and four ways to place the remaining teacher in one of the four spaces between successive students. Hence, there are $$3 \cdot 2 \cdot 4 \cdot 5!$$ linear arrangements in which teachers are at both ends of the row.
Hence, there are $$6 \cdot 5 \cdot 4 \cdot 5! - 3 \cdot 2 \cdot 4 \cdot 5! = (120 - 24)5! = 96 \cdot 5!$$ linear arrangements of teachers and students so that no two teachers are consecutive and teachers are not at both ends of the row.
These linear arrangements correspond to the permissible ways we can seat the students and teachers around the table. To account for rotational invariance, we divide the number of linear arrangements by $8$, which yields
$$\frac{96 \cdot 5!}{8} = 12 \cdot 5! = 1440$$
permissible seating arrangements around a circular table, as you found.
Best Answer
The numbers are counting different things. When you count the number of ways in which $n$ things may be distributed amongst $k$ ‘containers’, the answer depends on whether or not the things are distinguishable from one another, and it depends on whether or not the ‘containers’ are distinguishable from one another. It also depends on whether you require each ‘container’ to receive at least one of the objects.
The number of ways to distribute $10$ distinguishable teachers amongst $5$ distinguishable schools is $5^{10}$, not $10^5$. Teachers and schools are not usually considered to be indistinguishable, so this is the most reasonable answer. In the unlikely event that the teachers (but not the schools) are indistinguishable, there are $\binom{14}{4} = 1001$ ways to distribute them, not $\binom94 = 126$; $\binom94$ is the number of ways to distribute $10$ indistinguishable teachers amongst $5$ schools if each school is required to get at least one teacher. Thus, even assuming that $10^5$ is a careless error for $5^{10}$, the two answers offered in the question correspond to markedly different questions.
If we impose the boldface requirement on the version in which the teachers are distinguishable, the calculation is a bit more complicated. A standard inclusion-exclusion argument yields $$\begin{align*} \sum\limits_{k=0}^5 (-1)^k \binom{5}{k}(5-k)^{10} &= 5^{10} - \binom51 4^{10} + \binom52 3^{10} - \binom53 2^{10} + \binom54 1^{10}\\ &= 5^{10} - 5\cdot 4^{10} + 10\cdot 3^{10} - 10\cdot 2^{10} + 5\\ &=5,103,000, \end{align*}$$ significantly fewer than $5^{10} = 9,765,625$.
All of these calculations assume that the schools are distinguishable. If the teachers are distinguishable and the schools are not, and each school is required to get at least one teacher, the answer is $\left\{\begin{matrix}10\\5\end{matrix} \right\} = 42,525$, a Stirling number of the second kind. If one or more of the schools may receive no teacher, the answer is $$\sum\limits_{k=1}^5 \left\{\begin{matrix}10\\k\end{matrix}\right\} = 1+511+9330+34,105+42,525 = 86,472.$$
Finally, if neither teachers nor schools are distinguishable, we’re simply counting the number of partitions of $10$ into exactly $5$ or at most $5$ parts, depending on whether or not each school is to receive at least one teacher. These numbers are small enough to be calculated by direct enumeration. There is one partition of $10$ into one part. There are $5$ partitions of $10$ into $2$ parts: $$\begin{matrix}1+9; & 2+8; & 3+7; & 4+6; &5+5\end{matrix}$$. There are $8$ partitions of $10$ into $3$ parts: $$\begin{matrix}1+1+8; & 1+2+7; & 1+3+6; & 1+4+5\\ 2+2+6; & 2+3+5; & 2+4+4; & 3+3+4\end{matrix}$$ There are $9$ partitions of $10$ into $4$ parts: $$\begin{matrix}1+1+1+7; & 1+1+2+6; & 1+1+3+5\\ 1+1+4+4; & 1+2+2+5; & 1+2+3+4\\ 1+3+3+3; & 2+2+2+4; & 2+2+3+3\end{matrix}$$ And there are $4$ partitions of $10$ into $5$ parts: $$\begin{matrix}1+1+1+1+6; & 1+1+1+2+5\\ 1+1+1+3+4; & 2+2+2+2+2\end{matrix}$$
Thus, if neither schools nor teachers are indistinguishable, there are $27$ or $4$ ways to distribute the teachers, depending on whether or not every school must get at least one.