Why does a spring mass system oscillate? For simply harmonic motion,
acceleration $= -\omega^2 x$, where $\omega$ is the angular frequency.
Within limits of Hooke's law,
the restoring force on the spring is given by
$$F= -k \cdot x$$
This force fits the simple harmonic motion condition with $k=\omega^2m$.
But, we have $F = 0$ for $x=0$.
If I displace block (of some mass) attached to a spring (massless) rightward (on horizontal plane), and then release it, it would accelerate leftward because of above spring force.
But at the moment when $x=0$,  $F=0$ and thus $a=0$.
Then why does the spring block system experience simple harmonic motion?
Or why would the block accelerate farther leftward if there is no force on it?
I'm confused.
 A: 
Or why would the block accelerate farther leftward if there is no force on it?

Assuming the spring-mass system is horizontal then you are right: at $x = 0$ there is no net force on the block which means net acceleration is also zero. But you are forgetting that the block (mass) has certain velocity at that point and it keeps moving until negative acceleration provided by the spring restoring force stops it completely.

Then why does the spring block system experience simple harmonic motion?

Because of inertia of the block (mass) connected to the spring.

For simply harmonic motion, acceleration $= -\omega^2 x$, where $\omega$ is the angular frequency.

This is correct. From $F = -kx$ it follows that acceleration depends on the spring elongation $x$ which is defined as a sine function (harmonic oscillation). Therefore, the acceleration of the block itself is also a sine function.
A: But at the moment when $x=0,\, F=0$ and thus $a=0$, but the velocity $v\ne 0$ so the block overshoots the $x=0$ position and then has a force acting on it in the opposite direction to its direction of travel.
