Let's decouple, a bit, the iteration ranges (e.g. for loop control variable k) and what they are iterating over.
Technically, the loop control variable(s) iterate over a range of integers, e.g. 1..m or 1..n (rather than over elements of the arrays).
So, a loop, via its loop control variable, nominally iterates over a range of integers, not necessarily elements of an array, though it can be used to iterate through array elements, within the body of the loop, by using subscripting/indexing expressions like X(k) and Z(k) in terms of the loop control variable.
What exactly does k represent?
There is no reason that the question has to have used the same variable name, k, for both for-loops — but because it does use k twice (rather than, say, k1 first loop and k2 later), k means two different things in two different places. In Loop 1 k is an index ranging from 1 to m, which is the range of X, and in Loop 2, k is an index ranging from 1 to n, which is the range of Y.
m is the number of elements in the array (the dimension)
Specifically, in the array X — whereas n is the number in Y.
You already realize that the concatenation operation is concatenating X with Y not vice versa. Loop 1 then merely copies X into (the first part of) Z, character by character, and thus at each k, a copy from X into the same position in Z.
Next, Loop 2 copies Y into Z — but where to place the elements in Z, given that X has already been copied into Z?
In words, the answer is that the copy of (concatenation of) Y should go after the copy of X in Z.
In variables, that translates such that the first element of Y should go to the Z position at m because m is (firstly) the length of X, and also (secondly), it is where Loop 1 left off/stopped in Z — in other words, just after loop 1 finishes, m identifies the boundary between the last position used and the first free position in Z. Since we want to copy the whole of Y into Z, then all the characters of Y come from k but go to m + k in Z.
So, to recap, m is not only the size of X, but also where Loop 1 finished copying into Z. n is merely the size of Y. Note that X's copy in Z after Loop 1 now occupies Z(1..m), so then, the location where we want Y(1) to go is Z(m + 1), Y(2) to Z(m + 2), and so on, so that characters of Y are all placed after the entire copy of X in Z.
The copy of X in Z occupies Z(1..m) and the copy of Y in Z occupies Z(m..m+n), which is saying that the copy of X's comes first and the copy of Y comes directly after the copy of X.
(If you work through a real example, e.g. concatenate "hi " with "world", you'll see how this comes together.)
