Solution:
First, a lemma: For any $a$ with $0 \leq a \leq 2n$, $\binom{a}{k} + \binom{2n-a}{k} \geq 2\binom{n}{k}$. (By convention, we set $\binom{a}{k} = 0$ when $a < k$.)

Proof: We may assume $a \geq n$, since otherwise we can replace $a$ with $2n-a$. Now we prove the result by induction on $a$. For the base case, if $a = n$, then the lemma states that $2\binom{n}{k} \geq 2\binom{n}{k}$, which is trivial. If the lemma holds for some $a > 0$, then by the familiar identity,
$$\begin{aligned} & \left[\binom{a+1}{k} + \binom{2n-a-1}{k}\right] - \left[\binom{a}{k} + \binom{2n-a}{k}\right] \\ =\ & \left[\binom{a+1}{k} - \binom{a}{k}\right] - \left[\binom{2n-a}{k} - \binom{2n-a-1}{k}\right] \\ =\ & \binom{a}{k-1} - \binom{2n-a-1}{k-1} > 0 \end{aligned}$$
(since $a > 2n-a-1$), so $\binom{a+1}{k} + \binom{2n-a-1}{k} > \binom{a}{k} + \binom{2n-a}{k} \geq 2\binom{n}{k}$, giving the induction step. The lemma follows.

Now suppose that the elements of $A$ are labeled such that $a$ elements of the set $A$ receive the number 1 and $2n-a$ elements receive the number 2. Then the $k$-configuration can include all $k$-element subsets of $A$ except those contained among the $a$ elements numbered 1 or the $2n-a$ elements numbered 2. Thus, we have at most $\binom{2n}{k} - \binom{a}{k} - \binom{2n-a}{k}$ elements in the $k$-configuration, and by the lemma, this is at most
$$\binom{2n}{k} - 2\binom{n}{k}$$
On the other hand, we can achieve $\binom{2n}{k} - 2\binom{n}{k}$ via the recipe above—take all the $k$-element subsets of $A$, except those contained entirely within the first $n$ elements or entirely within the last $n$ elements. Then, labeling the first $n$ elements with the number 1 and the last $n$ elements with the number 2 shows that the configuration is 2-separable. So,
$$B_{k}(n) = \binom{2n}{k} - 2\binom{n}{k}$$