We want to have that $ f(f(x))=x $ for every $ x $. Since $ f(f(2))=2, $ we know $ f $ is its own inverse at $ x=2, $ so we can restrict our attention to $ x\neq 2 $. Since $ f $ applied to any number less than $ 2 $ returns a number greater than $ 2, $ and we can get all numbers greater than $ 2 $ this way, applying $ f $ to any number greater than $ 2 $ must give a number less than $ 2 $. Therefore $ k(x)<2 $ for any $ x>2 $. If $ x>2 $ and $ f $ is its own inverse then \[x=f(f(x))=f(k(x))=2+\left(k(x)-2\right)^2,\]where in the last step we used that $ k(x)<2 $. Subtracting $ 2 $ from both sides gives \[\left(k(x) - 2\right)^2 = x-2.\]Next, we recall that we must have $ k(x) < 2, $ so $ k(x) - 2 $ must be the negative number whose square is $ x-2 $. That is, we have $ k(x) - 2 = -\sqrt{x-2} $. Solving this for $ k(x) $ gives \[k(x)=\boxed{-\sqrt{x-2}+2}.\]