Given an initial ascending path contour, let $d_1, d_2, \dots, d_k$ denote the lengths of the horizontal segments, then $d_i > 0$ and $\sum_i d_i = N$. Because the contour starts in horizontal direction and has exactly $k$ horizontal and $k$ vertical segments, it ends with a vertical segment and so no horizontal segment is placed at level $N$. The lengths of the vertical segments is not related to the horizontal lengths and play no role in how the game is played so we will disregard them.

There are a total of $\binom{N-1}{k-1}$ ways to choose the lengths of the horizontal segments of an initial ascending path contour in an $(N, k)$-game. Indeed, we can encode the choice $(d_1, d_2, \dots, d_k)$ by a length $(N-1)$ sequence of 0-s and 1-s, with a 1 following each string of $(d_i - 1)$ consecutive 0-s to mark the end of the $i$-th segment, for $i \in \{1, \dots, k-1\}$ (the end of the last segment necessarily is at position $N$).

Depending on its position, a rock emerging on top of the $i$-th horizontal segment will require $0, 1, \dots, d_i - 2$ or $d_i - 1$ moves to be brought to a winning position. This gives a total of
$$0 + 1 + \dots + (d_i - 1) = \frac{(d_i - 1)d_i}{2} = \binom{d_i}{2}$$
moves for the $d_i$ possible positions of the rock above the $i$-th horizontal segment. Hence the average number of moves required for all $\binom{N-1}{k-1}$ initial paths and all $N$ possible starting positions of the rock is:
$$\sum_{(d_1, \dots, d_k)} \sum_{i=1}^{k} \binom{d_i}{2} / N \binom{N-1}{k-1}$$
We note that $\binom{d_i}{2}$ can also be interpreted as the number of ways to choose 2 distinct integer points on the $i$-th horizontal segment (but not including the starting point of the segment since then we would have a choice of $d_i + 1$ points). This is equivalent to splitting the $i$-th segment into 3 segments of lengths $a_i + b_i + c_i = d_i$, with $a_i > 0$ and $b_i > 0$ but possibly $c_i = 0$. For symmetry, let $c'_i = c_i + 1$. Then after relabelling the sequence $d_1, \dots, d_{i-1}, a_i, b_i, c'_i, d_{i+1}, \dots, d_k$ as $e_1, \dots, e_{k+2}$ and remembering $i \in \{1, \dots, k\}$ we get
$$\sum_{(d_1, \dots, d_k)} \sum_{i=1}^{k} \binom{d_i}{2} = k \cdot \# \left\{ (e_1, \dots, e_{k+2}) \middle| \sum_i e_i = N+1 \text{ and } e_i \in \mathbb{Z}, e_i > 0 \right\}$$
which is equal to $k \binom{N}{k+1}$ by binary code counting as before. Finally we get
$$\frac{k \binom{N}{k+1}}{N \binom{N-1}{k-1}} = \frac{N-k}{k+1},$$
as required.