Number the stations $1,2,\dots, N$ with $j,j+1$ being adjacent.
You need to pick $K$ numbers $x_1 \lt x_2 \lt \dots \lt x_K$ from these so that no two are consecutive.
The standard technique is to consider $y_i = x_i - (i-1)$.
These $y_i$ are distinct numbers in $1, 2, \dots, N-K+1$ and there is a $1-1$ mapping between the $x_i$ and $y_i$.
To get the $y_i$, we can pick $K$ distinct numbers from $1,2, \dots, N-K+1$ and just sort them in ascending order. Since this is again a $1-1$ mapping, the number we need is same as choosing $K$ distinct numbers from $1,2,\dots, N-K+1$ and thus the answer is
$$ \binom{N-K+1}{K}$$
The easiest way to solve it is to think of the four stops as fixed points and ask in how many ways the other stations can be inserted between them, if you must have at least one station between stops.
A | x x x | x x | x | x x B
Here, I’ve used bars to mark the four stops and $x$’s to mark the other stations. This is almost a standard stars-and-bars problem. You have five ‘blocks’ of stations, one before the first stop, three between stops, and one after the fourth stop. In my diagram I’ve put $0,3,2,1$, and $2$ stations into those blocks. Each of the middle three blocks must contain at least one station; the end blocks can be empty. In other words, I’m starting from this picture and have five stations left to distribute, each of which may go into any of the blocks numbered $1$ through $5$:
A | x | x | x | B
1 2 3 4 5
This amounts to counting the arrangements of a string of four bars and five $x$’s: there are $\binom94=126$ to choose which $4$ of the $9$ positions will be occupied by the bars.
Your approach will work if you count carefully enough. As you say, there are $6$ possibilities if the last three stops are at $8,10$, and $12$. Now leave the last two stops at $10$ and $12$ and move the second stop forward to $7$: there are just $5$ possibilities for the first stop. If you move the second stop forward to $6$, there are $4$, and so on, with just one when you move it all the way to $3$; these cases give you a total of $6+5+\ldots+1=\frac{6\cdot7}2=21$ possibilities. Now move the third stop forward to $9$; the same analysis will show that you have $5+\ldots+1=\frac{5\cdot6}2=15$ possibilites. As you keep moving the third stop forward, you get $4+3+2+1=10$, $3+2+1=6$, $2+1=3$, and $1$ possibility, for a total at this point of $21+15+10+6+3+1=56$ possibilities. Now repeat with the fourth stop moved forward to $11$. You’ll quickly find that most of the counting is a repetition of what you’ve already done, and you end up with $15+10+6+3+1=35$ possibilities. Similarly, with the last stop at $10$ you end up with $10+6+3+1=20$ possibilities. In the end you find yourself adding up $56+35+20+10+4+1$ to get $126$.
Best Answer
The idea you used to get the $35=\binom{7}{4}$ can be adapted to the circular case.
Cut the railway line just before a station, and straighten it out. Now we have $10$ stops in a line. Either (i) we use neither of the two endstations, $\binom{5}{4}$ ways; or else (ii) we use one endstation and therefore not the other.
If we select say the left endstation, then the next is forbidden, as is the other end. So we need to choose $3$ stops from $7$. This can be done in $\binom{5}{3}$ ways.
That gives a total of $\binom{5}{4}+2\binom{5}{3}$. The argument generalizes.