Let's go through each option to determine which one is FALSE.

The function is given as:

\[ f(x) = \begin{cases} -1, & \text{for } -2 \leq x \leq 0 \\ x - 1, & \text{for } 0 \leq x \leq 2 \end{cases} \]
Option A: \( f(|x|) = |x| - 1 \) for \( 0 \leq x \leq 2 \)

For \( |x| = x \) when \( x \geq 0 \), \( f(x) = x - 1 \), so \( f(|x|) = x - 1 \), which is equal to \( |x| - 1 \).
Hence, option A is true.

Option B: \( f(|x|) = x - 1 \) for \( 1 \leq x \leq 2 \)
For \( x \in [1, 2] \), \( f(x) = x - 1 \). Since \( |x| = x \) for positive \( x \), we have \( f(|x|) = x - 1 \), which is correct for this range.
Hence, option B is true.

Option C: \( f(|x|) + |f(x)| = 1 \) for \( 0 \leq x \leq 1 \)
For \( 0 \leq x \leq 1 \), \( f(x) = x - 1 \). Therefore, \( |f(x)| = |x - 1| = 1 - x \) (since \( x - 1 \) is negative or zero).
For \( |x| = x \), we have \( f(|x|) = x - 1 \). Thus: \[ f(|x|) + |f(x)| = (x - 1) + (1 - x) = 0. \] This does not equal 1.
Hence, option C is FALSE.

Option D: \( f(|x|) - |f(x)| = 1 \) for \( 1 \leq x \leq 2 \)
For \( 1 \leq x \leq 2 \), \( f(x) = x - 1 \), so \( |f(x)| = x - 1 \).
Also, for \( |x| = x \), we have \( f(|x|) = x - 1 \). Thus: \[ f(|x|) - |f(x)| = (x - 1) - (x - 1) = 0. \] This does not equal 1. - Hence, option D is also FALSE.

Conclusion: The false statement is C and D