I believe an explicit construction can be given as follows:
Note that all we need to do is find a bijection $\displaystyle F$ from $\displaystyle k$-sets to $\displaystyle k$-sets such that $\displaystyle F(X) \cap X = \emptyset$.
Given such an $\displaystyle F$, we can construct the required bijection $\displaystyle f$ by taking $\displaystyle f(X) = [n] - F(X)$, where $\displaystyle [n] = \{1,2,\dots, n\}$ and we are constructing subsets of $\displaystyle [n]$.
Now to construct the bijection $\displaystyle F$. Write $\displaystyle 1,2,\dots n$ along a circle.
Now given a set $\displaystyle X = \{x_1, x_2, \dots, x_k\}$ where $\displaystyle x_1 \lt x_2 \lt \dots x_k$.
Start at $\displaystyle x_1$ and go along the circle clockwise, picking the first element not in the set $\displaystyle X$ (kind of like consistent hashing, if you are a programmer). Then go to $\displaystyle x_2$ and go along clockwise to pick the the first element not picked so far (and not in $\displaystyle X$) and so on.
These $\displaystyle k$ elements you picked form $\displaystyle F(X)$. It is easy to see that $\displaystyle X \cap F(x) = \emptyset$.
We can also show that $\displaystyle F(X) \neq F(Y)$ if $X \neq Y$.
If $\displaystyle X = \{x_1, x_2, \dots, x_k\}$ and $x_1 \lt x_2 \dots$ and
if $\displaystyle Y = \{y_1, y_2, \dots, y_k\}$ and $y_1 \lt y_2 \dots$
Then suppose $j \gt 1$ is the smallest such that $\displaystyle x_j \neq y_j$ and assume $\displaystyle y_j \lt x_j$.
Now the above algorithm will either have picked $\displaystyle y_j$ to be in $\displaystyle F(X)$ or it would pick different numbers (for $\displaystyle F(X)$ and $\displaystyle F(Y)$) in the $\displaystyle j^{th}$ round.
If there is no such $\displaystyle j \gt 1$, then the sets only differ in the smallest element, and in which case, the resulting $\displaystyle F$ sets would be different too.
If we weren't worried about the exact construction, we can also give an existence proof.
The existence of a bijection can be shown by the following theorem:
Theorem: Every $\displaystyle k$-regular bipartite graph has a perfect matching.
This can be proven by induction on $\displaystyle k$ and using Hall's theorem.
You construct a bipartite graph with $\displaystyle k$-sets as the left partition and $\displaystyle n-k$-sets as the right partition.
Vertex $\displaystyle L_i$ is connected to $\displaystyle R_j$ iff $\displaystyle L_i \subset R_j$.
This is a $\displaystyle \binom{n-k}{n-2k}$ regular bipartite graph.