Confused with voltage doubler circuit
Let us analyze this circuit but we will freeze the time. 
I assumed \$V_p = 10V\$. 

For positive input from 0° to 90°, the situation is clear for you D2 is OFF. 
And will look like this:

 

Now let us see what will happen between 90° and 180° when \$V_p = 8.5V\$.



As you can see the D2 diode will conduct current because C2 was initially empty. 

But this will only happen at startup (when C2 is empty). 
Not during normal(steady-state) operation.   

And your book shows you the "steady-state" situation.