This is a problem about Diophantine equation.

The problem is the following.

If $ax+by=c$ is solvable and $b\ne0$, then prove that it has a solution $x_0$, $y_0$ with $0 \le x_0 <|b|$

First I thought that $x=x_1-\frac bg k$ , $y=y_1+\frac ag k$, where $g=gcd(a,b)$, $k$ is integer and $x_1$, $y_1$ is initial solution.

But, after this step, I am stuck and can't find what to do next.

You are nearly finished. Let us use a weaker result than the one you quoted. If $(x_1,y_1)$ is a solution, so is $(x_1-bk,y_1+ak)$ for any integer $k$. If $a$ and $b$ are not relatively prime, this does not produce all solutions, but it is enough for our needs.
By the result often called the Division Algorithm, there is a unique $r$, with $0\le r\lt |b|$, and an integer $q$, such that $x_1=q|b|+r$. This $r$ is the $x_0$ that we want. We pick $k=q$ if $b$ is positive, and $k=-q$ if $b$ is negative.