Consider paths from $(0,0)$ to $(k,n)$ that increase one of the two components on each step, so they make $n+k$ steps, among which $k$ horizontal (increasing $i$ in $(i,j)$) and $n$ vertical (increasing $j$).

There are $\binom{n+k}k$ such paths.

Let the first point with first coordinate $i$ be $(i,j_i)$ for $i=1,\ldots,k$, then clearly $0\leq j_1\leq\cdots\leq j_k\leq n$. Moreover all such sequences of values are allowed, and they determine the path: the horizontal steps are precisely those from $(i-1,j_i)$ to $(i,j_i)$ for $i=1,\ldots,k$.

Other points of view. If you take $y_i$ to be the index of the horizontal step to $(i,j_i)$ among all steps (which are numbered from $1$ to $n+k$), you'll get your $0<y_1<\cdots<y_k\leq n+k$. If you put a candy (star) at each point $(i,j)$ of the path and think of each horizontal step as a separation (bar), you'll have split your $n+k+1$ candies into $k+1$ vertical groups (for $i=0,1,\ldots,k$) the last group $i=k$ being the left-overs. But actually you want each group to have as many candies as it has vertical *steps*, not points, so you remove one from each group to correct this fencepost error, leaving groups of vertical steps of sizes $d_0,d_1,\ldots,d_k$ (each $d_i\geq0$), with $d_0+\cdots+d_k=n$. In the end you might just as well take $n+k$ steps, choose $k$ of them to be horizontal (bars) and the remaining $n$ to be vertical (stars). But I think I already said something similar at the beginning.