If
$f(x) = (x+2)^2+2$
then we get 2 inverse functions for this function:
If we input functions and inverse functions in to each other we must get "x" back. But why we don't get "x" when we input function $f(x) = (x+2)^2+2$ into second inverse function $f^{-1}(x) = -2-\sqrt{x-2}$
$f^-1(f(x)) = -2-\sqrt{f(x)-2}$
= $-2-\sqrt{(x+2)^2+2-2}$
= $-2-\sqrt{(x+2)^2}$
= $-2-(x+2)$
= $-2-x-2$
= $-4-x$ ≠ x
On
As a real-valued function on the reals $f$ is not an invertible function. So, treating some other function as if it were the inverse to $f$ is a recipe for disappointment.
A restriction of $f$, however, is an invertible function from $[-2, \infty)$ to $[2, \infty)$, and the first function you listed is the inverse.
Another restriction of $f$ is an invertible function from $(-\infty, -2]$ to $[2, \infty)$, and the second function you listed is the inverse.
In the second case, the error in your calculation of $f^{-1}(f(x))$ is that if $x \in (-\infty,- 2]$, then $\sqrt{(x+2)^2} \neq x+2$, since $\sqrt{}$ is being used to mean the positive square root, but $x+2$ is negative.
On
As you know the inverse has two branches. That's because $f(x)$ gives the same output for $x \in (-2,\infty)$ and $x \in (-\infty,-2)$
For $x > -2$, we take the positive square root as the inverse
$$ f^{-1}(f(x)) = -2 + \sqrt{f(x)-2} = -2 + |x+2| = -2 + (x+2) = x $$
For $x < -2$, we take the other square root
$$ f^{-1}(f(x)) = -2 - \sqrt{f(x)-2} = -2 - |x+2| = -2 + (x+2) = x $$
Note that $$\sqrt{(x+2)^2} = |x+2| = \begin{cases} x + 2, & x > -2 \\ -(x+2), & x < -2 \end{cases} $$
The absolute value changes the sign, which is why your simplification for the negative root is incorrect.
We have $$f(-\sqrt{x-2}-2+2)^2-2=x-2+2=x$$