Does $1$ or $2$ come first in the sequence $i + k, i + 2k, i + 3k...$ modulo $n$? Given coprime numbers $k$ and $n$, we can take any number $i$ and keep adding $k$ to it until we reach a number which is congruent to either $1$ or $2$ modulo $n$. This classifies the numbers between $1$ and $n$ into two classes - the ones which "hit" $1$ before $2$, and the ones that "hit" $2$ before $1$.
What can be said about these two classes of numbers?

I'm not sure that this is what you wanted to study, but here comes anyway.
Let $\ell$ be the multiplicative inverse of $k$, so $\ell k\equiv1\pmod n$ and $0<\ell<n$.
Consider the arithmetic sequence of elements of $\Bbb{Z}_n$ (with increment $k$)
$$1+k,1+2k,1+3k,\ldots,1+\ell k\equiv2,2+k,2+2k,\ldots,1-k,1$$
that repeats itself in a period of length $n$. From this sequence it is clear that if you begin at any of $\ell$ first numbers you hit $2$ first. OTOH, if you begin from any of the other numbers you hit $1$ first. The point is that no matter where you start from, you will always follow a cyclic shift of this sequence.

Example: Let $n=14$, $k=5$. Because $3\cdot5=15\equiv1\pmod{14}$ we have $\ell=3$. One period of the above sequence then reads (increment by $5$, the second period greyed out)
$$
6,11,2,7,12,3,8,13,4,9,0,5,10,1\color{grey}{(,6,11,2,\ldots)}.
$$
So if we start from $6,11$ or $2$ we hit $2$ first. That is $\ell=3$ cases as promised. In the remaining $14-\ell=11$ cases we hit $1$ first.
So altogether from $\ell$ starting points the sequence hits $2$ first and from $n-\ell$ starting points the sequence hit $1$ first. 
Observe that you get this same $\ell$ vs. $n-\ell$ split for all pairs $(a,a+1)$ in place of $(1,2)$.
