Generally speaking, you want to write absolute values of functions piecewise to get your hands dirty and solve for critical points. This means we have to find the sign of the function between zeros and apply the definition of absolute value.
As you noted, there are zeros at 0 and 2; these are the only two zeros of $x^2-2x$. We can check signs between and outside these points to find that
$x^2-2x>0$ if $x>2$ or $x<0$
$x^2-2x<0$ if $0
Therefore our function $f$ may be written
$f(x)=\begin{cases}x^2-2x & \text{if } x\leq 0 \\ 2x-x^2 & \text{if } 0\leq x\leq 2 \\ x^2-2x & \text{if } 2\leq x \end{cases}$
To find critical points, we must find points where the derivative does not exist or equals zero. In this case, it suffices to differentiate each piece; if the derivatives do not agree at the endpoints, then the derivative does not exist at the endpoint. $f'(x)=\begin{cases}2x-2 & \text{if } x< 0 \\ \text{undefined} & \text{if } x= 0 \\ 2-2x & \text{if } 0< x\leq 2 \\ \text{undefined} & \text{if } x= 2 \\ 2x-2 & \text{if } 2\leq x \end{cases}$
So we see that the derivative is undefined at $0$ and $2$. It is defined everywhere else, but it is easy to check the only point where the derivative is zero is when $x=1$.
(By the way, the comparison of the derivatives to the left and right of $0$, $2$ is the same as evaluating the left-hand derivative and right-hand derivative).
(EDIT: The definition of critical point may vary in what cases we include, but many standard calculus texts include non-existence in the definition for optimization.)