A question on vertical spring mass systems 
For a given mass attached to a vertical spring (near the surface of the earth), how do we know that the equilibrium point is halfway between its oscillation? (Let the height at the bottom of oscillation = 0.)

What I know:
At the equilibrium point, acceleration = 0, so $mg=kx$, where $x$ is the displacement between point of release and point of equilibrium.
How can I move on from here?
 A: If the mass was displaced a further distance $dx$ so the extension is $x+\delta x$ then the restoring force is $F = k(x+\delta x) - mg = kx + k \delta x -kx = k\delta x$
towards the equilibrium position.
The restoring force is proportional to the distance from the equilibrium position and that's the condition for SHM.
In SHM the extremes of the oscillation are equally spaced either side of the equilibrium position.
A: Overall you have made the problem more difficult by making the statement Let the height at the bottom of oscillation = 0.  Choosing the datum as the equilibrium position is the more usual way.
In the way you have formulated the question the diagram below shows: the unextended spring (natural length), the system (spring & mass) at equilibrium, the system when the mass is highest (Up) and the system when the mass is lowest (Down).

Equation energies at the very top and the very bottom gives,  $mg(y+y_0) + \frac 12 k (x-Y)^2 = \dfrac 12 k (x+y_0)^2$ which together with $mg = kx$ will produce the result that $Y=y_0$ ie the motion is symmetrical about the equilibrium position.
