We can systematically decompose the problem top-down. Let $f(n)$ be the direction in which the $n$-th person (with indices wrapping round the ring) does his disgusting thing, $-1,0,1$ for left, centre and right respectively. Note that the number of ways is equivalent to the number of possible $f$ that satisfies the corresponding conditions, as long as $n \ge 3$. The obvious way is to take out one person. If $f(0) = 0$, then we need to know the number of ways for $(n-1)$ people such that $f(0) \ne 1$ and $f(1) \ne -1$. If $f(0) = 1$, then either $f(1) = -1$ or $f(-1) = 1$. In the first case we need to know the number of ways for $(n-2)$ people such that $f(0) \ne 1$ and $f(1) \ne -1$. In the second case we need to know the number of ways for $(n-1)$ people such that $f(0) = 1$ and $f(1) \ne -1$. I will leave you to verify the bijection in each of these reductions. These reductions tell us to define the following:
$\def\nn{\mathbb{N}}$
Let $a(n) = \#(\text{ways for $n$ people such that $f(0) \ne 1$ and $f(1) \ne -1$})$.
Let $b(n) = \#(\text{ways for $n$ people such that $f(0) = 1$ and $f(1) \ne -1$})$.
Then $a(n+2) = a(n+1) + a(n)$ for any $n \in \nn^+$. [Either $f(1) = 0$ or $(f(1),f(2)) = (1,-1)$.]
Also $b(n+1) = b(n)$ for any $n \in \nn^+$. [$f(-1) = 1$.]
You can check that $a(1) = 1$ and $a(2) = 2$, and hence $a(n) = F_{n+1}$ for any $n \in \nn^+$ where $F_n$ is the $n$-th Fibonacci number. You can also check that $b(1) = 1$, and hence $b(n) = 1$ for any $n \in \nn^+$.
Therefore the total number of ways for $n$ people is
$a(n-1) + 2 a(n-2) + 2 b(n)$
$\ = F_n + 2 F_{n-1} + 2$
$\ = F_{n+1} + F_{n-1} + 2$.
Notes
This method gives exactly the same answer as Henry's method, but is more general and can be applied to other more complicated problems without requiring any insight.