Constructing a k-perfect permutations family
Here is a sketch on how to do almost what you want, but not quite. 
It only proves a $k^{k+\operatorname{polylog}(k)}\log n$ bound.

Consider a set of size $m=n\log_2 k$ which we think of as $n$ groups on $\log_2 k$ elements each.

Next construct an $(m,k\log_2 k)$-universal set family $F$ of size $2^{k\log k+\operatorname{polylog}(k)}\log m$ using 
the method of Naor, Schulman, and Srinivasan. We interprete each set in $F$ as a string on $n$ symbols from $[k]$ with one symbol for each group.

Now we turn each string into a permutation of $[n]$ in the following way:

We first think of the $n$ first positive integers divided in $k$ stacks $T_0$ throgh $T_{k-1}$ where $T_0$ contains the elements $1,...,n/k$ from top to bottom, $T_1$ contains $n/k+1,...,2n/k$, and so on.

We parse each string $s_1s_2...s_n\in F$ into a permutation by considering the symbols one at a time for increasing $j$, and on encountering symbol $s_j$ we pop the stack
$T_{s_j}$ from the top and put that number in place $j$ of the permutation.

Since the family $F$ was $(m,k\log_2 k)$-universal, we know that for every ordered set of $k$ of the $n$ groups, there is one string that maps the symbols $0$,$1$,...,$k-1$ to that ordered set. By the construction of the stacks, the elements fetched to the permutations preserve the order required.