Oscillation of particle in harmonic oscillator
Classically speaking, the (idealised) harmonic oscillator is defined by Hooke's law. Quoting from Wikipedia:

In classical mechanics, a harmonic oscillator is a system that, when displaced from its equilibrium position, experiences a restoring force $F$ proportional to the displacement $x$:
$$\vec{F} = -k\vec{x}$$
where $k$ is a positive constant. If $F$ is the only force acting on the system, the system is called a simple harmonic oscillator.

In one dimension we can drop the vectors and write this as $F = -kx$. But the force is also the (negative) derivative of the potential, i.e. $F = -\mathrm dV/\mathrm dx$, and integrating this to get $V = kx^2/2 + C$ is trivial. The constant of integration $C$ is arbitrary, but it doesn't fundamentally change the physics (it just moves the energy scale), so we can ditch it by setting it to zero.
If you had a potential energy with higher-order terms in $x$ (i.e. $x^3$ and above), then the derivative with respect to $x$ would contain terms that are $x^2$ and above. This directly contradicts the definition of the harmonic oscillator. It could be some other kind of oscillator, but its motion would not be called simple harmonic.
Consequently, for this particular oscillator, trying to perform a Taylor expansion is a little pointless: there are no terms in $x^3$ or above, so all the higher-order derivatives $\mathrm d^3V/\mathrm dx^3, \cdots$ are all zero, and all those terms in the Taylor expansion are zero. Of course, in the general case, Taylor analysis of a different oscillator could be very useful. Just not here.
The potential energy of the quantum harmonic oscillator is directly inherited from the classical simple harmonic oscillator, by replacing the classical quantities $x$ and $p~ (= mv)$ with their quantum analogues $\hat{x}$ and $\hat{p} ~(= -\mathrm i\hbar(\partial/\partial x))$. So given that the classical potential energy is $V = kx^2/2$, it's not too much of a stretch to get the quantum potential energy $V = kx^2/2$.