Meaning of$~\sum_\text{cyclic}$?

There are six permutations of the variables $(x, y, z)$, namely $(x, y, z)$, $(x, z, y)$, $(y, x, z)$, $(y, z, x)$, $(z, x, y)$, and $(z, y, x)$.

There are three cyclic permutations of the variables $(x, y, z)$, namely $(x, y, z)$, $(y, z, x)$, and $(z, x, y)$.

So the sum of the expression $(x - y)^z$ over all permutations is given by

$$(x - y)^z + (x - z)^y + (y - x)^z + (y - z)^x + (z - x)^y + (z - y)^x$$

whereas the sum of the expression $(x - y)^z$ over all cyclic permutations is given by

$$(x - y)^z + (y - z)^x + (z - x)^y.$$

Note that the two sums are not the same; the first contains more summands as there are more permutations than cyclic permutations (i.e. not every permutation is a cyclic permutation, e.g. $(x, z, y)$ is a permutation of $(x, y, z)$ but not a cyclic permutation).

If you are familiar with abstract algebra, this is a special case of a more general notion.

Suppose you have a finite group $G$ acting on a space $X$ and a function $f\colon X\to {\bf R}$ (here, you can put any abelian group). Then you can consider the summation over $G$: $\sum_{g\in G} f\circ g$. This is a special case where $X={\bf R}^3$ and $G={\bf Z}_3$ acts by cyclically permuting (hence the name) the coordinates, which are $x,y$ and $z$ in the question you linked. The different case the link refers to is if you had taken instead $G=S_3$ (the full permutation group).

Frequently, this operation is paired with averaging, so we take instead $\frac{1}{\lvert G\rvert}\sum_{g\in G} f\circ g$. Note that the result of this operation is invariant under the action of $G$ and can be considered a projection of $f$ onto the space of $G$-invariant functions (because it does not change $f$ it it was $G$-invariant to begin with).

More generally, if you have a possibly infinite group $G$ with a sufficiently nice finite invariant measure $\mu$, you can do the same trick by looking at $\frac{1}{\mu(G)}\int_{G}f\circ g\,\textrm{d}\mu(g)$, again yielding an invariant function. In case of finite groups this $\mu$ is just the counting measure. This kind of averaging is a very useful idea in a variety of contexts.