Unfortunately, I can't quite grasp where $i + nk$ is coming from, as I thought we were manipulating subscripts modulo $k$. If subscripts are computed modulo $k$, then the cycle $(x_i, x_{i + k}, \ldots, x_{i + (k-k)})$ must be the 1-cycle $(x_i)$, since $i \equiv_k i + k$. In this case, you can conclude that $\sigma^k = e$, without mentioning $i + nk$.
You had me up to there, but I'm just not sure how you bring it all together. You definitely figured out a lot about the structure of $\sigma^m$, which means there's a more direct route find the order of the $k$-cycle $\sigma$.
Personally, I would use the fact that your permutation $\sigma = (x_1 x_2 \ldots x_k)$ is in the symmetric group $S_n$, which acts on the set $[n] = \{1, 2, \ldots, n\}$.
So, we could show that if we apply the map $\sigma: [n] \to [n]$ exactly $k$ times to any $i \in [n]$, then we wind up back at $i$; that is, $\sigma^k(i) = i$, and that $\sigma^m \neq e$ when $m < k$ and $k > 1$.
For $i \notin \{x_1, \ldots, x_k\}$, $\sigma^k(i)$ is clearly $i$, since $\sigma(i) = i$.
Otherwise, we know that $\sigma(x_j) = x_{j+1}$, with subscripts computed mod $k$, as you pointed out. It follows that $\sigma^m(x_i) = x_{i + m}$, which will be $x_i$ only when $m = k$, since the subscripts are computed mod $k$ and $x_i \neq x_j$ whenever $i \not\equiv_k j$
If you decompose into cycles first, all you need to do is express each cycle as a product of transpositions. There are various ways to do this, for example
$$ (1\,2\,3\,4\,\ldots\,n) = (1\,n)\cdots(1\,4)(1\,3)(1\,2) $$
or
$$ (1\,2\,3\,4\,\ldots\,n) = (1\,2)(2\,3)(3\,4)\cdots(n{-}1\;n) $$
Best Answer
Yes, there are competing conventions for multiplication in the symmetric group.
But some people reverse the function notation, as well. Reader beware…
I prefer the second one, for the same reasons that Olivia outlines. But my understanding is that the first one is particularly common in combinatorial or geometric group theory.