First, ignore the condition that adjacent vertices cannot be paired up. The number of ways to pair up the vertices into $n/2$ disjoint pairs is $n!!$, the double factorial, defined by
$$
(n-1)!!\stackrel{\text{def}}=(n-1)(n-3)(n-5)\cdots=\prod_{i=0}^{\lfloor n/2\rfloor-1}(n-1-2i).
$$
To account for the condition that adjacent vertices cannot be paired, we use the principle of inclusion exclusion.
Let $S$ be the set of all vertex pairings, so $|S|=n!!$. For each $i\in \{1,\dots,n\}$, let $E_i$ be the set of vertex pairings where vertex number $i$ is paired with its clockwise neighbor. I assume the vertices so the clockwise neighbor of $i$ is $i+1$ (with addition wrapping modulo $n$). The set of legal pairings is $S\setminus (E_1\cup \dots \cup E_n)$, and PIE tells us that
$$
\begin{align}
|S\setminus (E_1\cup \dots \cup E_n)|
&=\sum_{k=0}^n(-1)^k\sum_{1\le i_1<\dots <i_k\le n}|E_{i_1}\cap \dots \cap E_{i_k}|.
\end{align}\tag1
$$
To evaluate $|E_{i_1}\cap \dots \cap E_{i_k}|$, there are two cases.
Suppose the list of indices $[i_1,\dots,i_k]$ contains an adjacent pair, meaning that there exists $r\in \{1,\dots,k\}$ such that $i_{r+1}=i_r+1$ (or, $i_1\equiv i_n+1\pmod n$). In this case, $|E_{i_1}\cap \dots \cap E_{i_k}|=0$. This is because it is impossible to simultaneously have vertex number $i_r$ paired with $i_r+1$, and to have vertex $i_{r}+1$ paired with $i_r+2$.
Otherwise, the indices represent pairwise nonadjacent vertices. In this case, there are $k$ particular vertices which must be paired with their clockwise neighbors. This leaves $n-2k$ vertices to be paired up arbitrarily, which can be done in $(n-1-2k)!!$ ways.
In summary, the innermost summation in $(1)$ has two types of terms: zeroes, and $(n-1-2k)!!$. Therefore, instead of summing over all ways to choose indices such that $1\le i_1<\dots <i_k\le n$, we can simply count the number of ways to choose these indices which fall in case $2$, and then multiply by $(n-1-2k)!!$. That is, if we let $N(n,k)$ be the number of ways to select $k$ nonadjacent vertices on an $n$-gon, then the final answer is
$$
|S\setminus (E_1\cup \dots \cup E_n)|=\sum_{k=0}^{\lfloor n/2\rfloor} (-1)^k N(n,k) (n-1-2k)!!
$$
Note: When $k=n/2$, the term is $N(n,n/2)· (-1)!!$. By definition, $(-1)!!=1$. This is the convention that makes the most sense, like how we define $0!$ to be $1$.
All that remains is to evaluate $N(n,k)$. There is a simple formula for $N(n,k)$ in terms of $n$ and $k$ that involves a binomial coefficient. It is a nice exercise to try to work it out, so I encourage you to do so. If you get stuck, see this old MSE question.
