Your explanation is difficult to follow when you do distributivity of % over +. I mean when you move from

    (i + t) % n = (j + 2t) % n

    (i % n) + (t % n) = (j % n) + (2t % n)

    (5 + x) % 3 = 4
    5 % 3 + x % 3 = 4

It's better to express your explanation using "= (mod n)" from the beginning.

    (i + t) % n = (j + 2t) % n 
    i + t = j + 2t (mod n)

For particular values of x, y, and loop period there may be values of i and j such that going forward from those they never meet, but it is not possible to get into one of those positions if a past state had i == j.

Proof:

In the following I'll use your notation for the most part, except allowing for a "lead in" of length L from that common starting node to the start of the loop, so the nodes are numbered 0, 1, 2, ..., L-1, L, L+1, ..., L+n-1, with the nodes in the loop being {L, L+1, L+2, ..., L+n-1}.

At time t, i has taken xt steps and j has taken yt steps. Let t >= L/x and t >= L/y, so that i and j are both past the lead in into the loop.

Then i is xt-L steps into the loop, and j is yt-L into the loop. These will be at the same spot in the loop if xt-L = yt-L mod n, which happens whenever (x-y)t = 0 mod n. All multiples of n are solutions of this, so just pick any that are >= L/x and L/y, and you have a solution.

If i and j do not start at the same place, all bets are off. That 0 is replaced with the differences between the starting node numbers, and as your example shows may not have a solution for some combinations of parameters.

Maybe it would be helpful to rewrite the loop with a condition as you described, and %= i and j with n after incrementing (by 1 or by 2) both variables in the loop body.