Let's summarize Jaap Scherphuis's answer to the easier version of the problem where the cat can only move 1 step at a time:
The idea is that if we know the cat started off somewhere in the first $n$ boxes, there is a finite sequence of moves (whose length depends on $n$) that's guaranteed to catch the cat. We start by guessing that the cat started in the first $9$ boxes, and try to catch it. If we don't get it, we next guess it started in the first $99$ boxes and try again; then the first $999$, $9999$, and so on. Eventually one of our guesses has to be right so we're guaranteed to catch the cat.
If we use the shorthand $S(n)$ to mean "perform a series of moves that will catch the cat if it started in the first $n$ boxes", then the strategy is: perform $S(9), S(99), S(999), ...$ etc. until we catch the cat. (Of course the sequence $9, 99, 999, ...$ is arbitrary and can be any sequence that grows to infinity.)
Now, since in this version the cat has a step size of $k$, we have to modify our strategy:
by just guessing what $k$ is! Just like in the 1-step case, if we know both the step size $k$ and a bound $n$ on the starting position of the cat, there's a finite sequence of moves we can make that's guaranteed to catch it. As an extension of the previous notation, let's denote that by $S_k(n)$, so the $S(n)$ from before is now $S_1(n)$. Then we proceed similarly to how we did before: first, guess that the step size is $1$, and try to catch the cat. If we're right, great! If not, guess the step size is $2$ and try again, then $3$, $4$, and so on till we get it.
...Right?
Not quite. This doesn't work because:
If we write out the sequences of moves that we're making, the first step (checking $k=1$) already involves an infinite sequence $S_1(9), S_1(99), S_1(999),...$ meaning that if $k\ne 1$, we'll make an infinite number of moves without finding the cat. Essentially, we're trying to make the moves in the following order:$$S_1(9),S_1(99),S_1(999),...\\S_2(9),S_2(99),S_2(999),...\\S_3(9),S_3(99),S_3(999),...\\\vdots$$ which doesn't work since there are an infinite number of $S_1$'s to try before the first $S_2$.
We can fix it by:
Interleaving the values of $n$ and $k$ that we try, instead of trying to exhaust the entirety of $k=1$ before proceeding to $k=2$. (Essentially, we need to solve the problem of enumerating ordered pairs of natural numbers.) For example, we can do our moves in the following order:\begin{align*}&S_1(9),\\&S_2(9),S_1(99),\\&S_3(9),S_2(99),S_1(999),\\&...\end{align*}This way no matter the actual value of $k$ and $n$, we're guaranteed to reach it after a finite number of moves.
Finally, here's an appendix of sorts to expand on some details:
What if the cat keeps moving farther away faster than we expand our search?
This isn't a problem because the $n$ in $S_k(n)$ is the starting position of the cat, not the current position. This means that $S_k(n)$ really depends not only on $k$ and $n$, but also on how many turns have already passed before we invoke it: if we're performing it after we already made $x$ moves, then it really means "assume the cat has a step size $k$ and started at below $n$, so it currently is at position at most $n+kx$." Although the number of steps this takes depends on $x$, not just $n$ and $k$, the important thing is that it's always finite.
What actually is the strategy $S_k(n)$?
For $k=1$, see the previous question's answer. For $k>1$, we can reduce it to the 1-step case by making another guess, this time guessing what the cat's position is mod $k$. For example, take $k=10$. If the cat started at a position ending in $7$, it can only move between positions $7, 17, 27, 37, ...$ which is just like the 1-step case with the boxes relabeled. We just have to check each possible value mod $k$, doing $k$ different shifted versions of the 1-step strategy.
