Take the breakdown voltage of the Zener diode used in the given circuit as 6V. For the input voltage shown in figure below, the time variation of the output voltage is :
(Graphs drawn are schematic and not to scale)

During each half cycle, when $V_{in}$  builds from 0 to 6V, one of the diodes is in reverse biased but not yet reached break down. 

So, the diode branch is open and bypasses the  $V_{in}$,  it will follow the input voltage.
Once  $V_{in}$ builds from 6V to 10V, then one Zener achieves breakdown and other in forward bias  $\left(\Delta V=0\right)$ produces constant voltage.