jmc

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 $x\ne 2.$

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 $$k(x)=\boxed{-\sqrt{x-2}+2}.$$