You have asked a good question.
…it does not seem to me that the projectile is simple parabola (as in the ground case) that can be "cut" into halves. So, why does this work?
Really it contrasts with the symmetry. But it has nothing to do with symmetry. You will better understand with following graphs.
Here black line indicates the trajectory of the ball and brown line is the inclined plane.
As @John Rennie precisely explained,
....the word "peak" means maximum distance from the plane measured normal to the plane. It does not mean the maximum height measured from the horizontal.
I have marked them as 'A' and 'B' respectively.
If you try graphing this yourself you will realize that distances $a$ and $b$ in the following picture are equal (This property was discovered by Archimedes).*
Since there is no horizontal acceleration, the amounts of time that the ball takes to pass $a$ and $b$ are equal (consider horizontal motion). Eventually you will conclude that that the time of the flight is given by doubling the time to reach the peak(A).
Thus it is not due to symmetry, but because of the special property of intersection of a straight line and a parabola.
Hope this helps.
P.S.: @Fredriksy has also explained the same thing in his answer,
It seems like you are worried that the flight path after you rotate the picture (x-y axes) is NOT a parabola. However this is not actually important for determining the time of flight.
I guess, with this explanation and my graphs you will understand better. Good luck.
*You can find the mathematical proof here.
(Special thanks go to @CiaPan and @Pope)
EDIT:
Can you observe something else interesting? If you consider a projection relative to the horizontal plane, the horizontal plane will also be a chord to the trajectory, which is a parabola. So the observation, 'by doubling the time it takes to reach maximum altitude with respect to the considered plane, flight time is obtained ', can also be interpreted as a result of this special property of intersection, although it obviously seems to be a consequence due to symmetry:-)
The extra height will give you more travel time before it strikes the ground, and hence more time to travel horizontally. In an ideal situation, the projectile maintains its horizontal speed until it hits the ground. If the ground is further "down" from where it was fired, it means it will travel further, all else being equal.
As far as the optimal launch angle in this situation, it will be less than 45° as that corresponds to the ideal angle for level ground.
Mathmaticall if the launch height is $h$ above ground, then the time for impact is
$$ t = \sqrt{ \left( \frac{v \sin \theta}{g}\right)^2 + \frac{2 h}{g} } + \frac{v \sin \theta}{g} $$
and so the range is
$$ x = v \cos\theta \sqrt{ \left( \frac{v \sin \theta}{g}\right)^2 + \frac{2 h}{g} } + \frac{v^2 \sin \theta \cos \theta}{g} $$
Now the optimal angle for launch maximizes the range, and it comes out to be (in degrees)
$$ \theta° = 90° - \arctan \left( \sqrt{1+ \frac{2 g h}{v^2}} \right) $$
You can see when $h=0$, the expression above is $\theta° = 90° - \arctan(1) = 45°$. For positive values of $h$ the angle becomes less than 45°.
Best Answer
It does not. The diagram is incorrect, as it assumes that the path becomes a straight line after it reaches $y = 100 \text{m}$ from the ground. The path still remains a parabola, so the final angle($\theta_f$) is different from the initial angle at which the bullet was shot ($\theta_i$)
To solve the question, find the range $R$ for any arbitrary $\theta_i$ and set $\frac{dR}{d\theta_i}$ to zero to maximise $R$ in terms of $\theta_i$.
Hope this helps.