Understanding why any permutation $\sigma \in S_n$ has a cyclid decomposition

Question

Consider the first theorem presented here which proves that any permutation $\sigma \in S_n$ has a cyclic decomposition. As I've understood from the given proof, it is stated that $\sigma \in S_n$ has a cyclic decomposition because the size of the orbit of a given $x$ is $d$, where $d$ is the order of the permutation $\sigma$. As saddening as it may sound, I'm completely missing the crux of the proof as I don't understand why the fact that the size of the orbit of some $x$ is equal to the size of the permutation $\sigma$ means that this cyclic representation of the orbit of $x$ "decomposes" the permutation $\sigma$. I.e. why do we have a cyclic decomposition for the arbitrary permutation $\sigma$?