Why does capacitor voltage lag current?
The picture in your question assumes that the voltage waveform started some time earlier and that the transient of it beginning is no longer affecting things.

Basically Q=CV and this translates to I = C dv/dt and, if you applied a sinewave the differential of that sinewave voltage gives rise to the cosine wave of current but, of course at t=0 things are a little different; For a start you can't suddenly start a sinewave from rest - that would imply infinite bandwidth. Given this fact, there is a small finite time which the current rapidly ramps up to the starting value in your picture. From thereon it pretty much follows the equation given above.



EDIT section, mechanical analogy

A mechancial analogy could be regarded as a flywheel i.e. a rotating mass. The force applied to the end of the flywheel will accelerate the speed at which the flywheel rotates but when the flywheel (lossless assumed) is at constant speed, no force is needed. You can imagine the flywheel speed like voltage; the flywheel has charged up to speed n and there is no longer any force needed to keep it charged at that speed. Just like a capacitor, once charged to a constant voltage there is no current needed to keep a perfect capacitor at that voltage.

However, if you applied a constant force to decelerate the flywheel, the speed decelerates linearly and if the constant force is a true constant force, the flywheel speed will decelerate through n=0 and start rotating in the opposite direction after a little while. Force is -X and speed ramps down linearly. Ditto with the capacitor, if you take a constant current from the capacitor the voltage falls linearly and eventually becomes negative and charges up to a negative voltage.