Maximal size of a set of ordered words such that no pair of letters occurs twice
There is a simple upper bound of $$\frac{\binom{n}{2}}{k-1},$$ following from the fact that there are $\binom{n}{2}$ ordered pairs of elements, and each ordered word contains $k-1$ of them.

There is an almost matching lower bound of
$$
\frac{\binom{n-k+1}{2}}{k-1}.
$$
This shows that for fixed $k$, the answer is asymptotically
$$
\frac{n^2}{2(k-1)} \pm O(n).
$$

For every $c \geq 1$ and $1 \leq d \leq c$, we can consider the collection of ordered words of the form
$$
d,d+c,\ldots,d+(k-1)c \\
d+(k-1)c,d+(k+1)c,\ldots,d+2(k-1)c \\
\ldots
$$
These collections for all $c,d$ have disjoint pairs. For a given $m \leq n$ and $c$, there is a word of this type if $m-(k-1)c \geq 1$, that is, if $c \leq \frac{m-1}{k-1}$. Therefore, the total number of words is
$$
\sum_{m=k}^n \left\lfloor \frac{m-1}{k-1} \right\rfloor.
$$
We can estimate this sum roughly by
$$
\sum_{m=k}^n \left(\frac{m-1}{k-1}-1\right) =
\sum_{m=k}^n \frac{m-k}{k-1} = \frac{\binom{n-k+1}{2}}{k-1}.
$$