Your understanding is basically right. A helpful way to understand the boundary where $\omega = \omega_p$ is to look at the refractive index,
\begin{equation}
n=\frac{c}{v_{p}}=\sqrt{\frac{\epsilon \mu}{\epsilon_0 \mu_0}}\approx\sqrt{\epsilon_r}\,,
\end{equation}
where $v_p$ is the phase velocity and I have taken $\mu_0\approx \mu$.
For electromagnetic (EM) waves incident on a cold unmagnetized plasma the ions can be assumed to be at rest relative to the electrons due to their inertia. The much lighter electrons move more readily in response to the electric field of an EM wave. The displacement $x$ between an electron and ion due to the EM wave creates an electric dipole moment, which affects the plasma's relative permittivity $\epsilon_r$,
\begin{equation}
\epsilon_r =1+\chi_e = 1 + \frac{P}{\epsilon_0 E}= 1 + \frac{n_e p_e}{\epsilon_0 E} =1+\frac{n_e ex}{\epsilon_0 E}\,,
\end{equation}
where $E= E_0 e^{-i\omega t}$ is the electric field of the EM wave, $P= n_e p_e$ is the polarization density (which aligns with the electric field in a simple medium), $n_e$ is the electron number density and $p_e =e x$ is the electric dipole moment.
The displacement $x$ is given by the equation of motion $\ddot{x} = \frac{e}{m_e} E$ and also follows simple harmonic motion as Sean has suggested, $\ddot{x} = -\omega^2 x$, so
\begin{equation}
x = -\frac{\ddot{x}}{\omega^2}= -\frac{e E}{m_e \omega^2}\,.
\end{equation}
Plugging this into our equation for the relative permittivity gives
\begin{equation}
\epsilon_r = 1 - \frac{n_e e^2}{\epsilon_0 m_e \omega^2}=1 - \frac{\omega_p^2}{\omega^2}\,.
\end{equation}
Therefore, the refractive index becomes zero at $\omega=\omega_p$ and is imaginary for $\omega<\omega_p$.
Now we consider the various cases depending on the frequency of the EM wave. Generally, in a cold plasma the electrons are able to respond to the EM wave on a characteristic timescale of $1/\omega_p$:
- For $\omega<\omega_p$, the math tells us that waves cannot propagate in a medium with imaginary refractive index (the solution to the wave equation falls off exponentially and the field becomes evanescent). Physically, the electrons are able to cancel the electric field of the EM wave within the period of oscillation of the wave ($\sim 1/\omega$). Thus, the cold plasma behaves like a metal, which has essentially free electrons that cancel fields tangential to the surface. You are therefore correct to say that the electrons establish a reflected field of radiation emanating from the plasma's surface. The majority of the wave energy bounces off the plasma and does not enter it (e.g. radio waves bouncing off ionosphere).
- For $\omega=\omega_p$, the wave energy gets absorbed by the plasma with little reflection.
- For $\omega>\omega_p$, the electrons are unable to completely cancel the electric field of the EM wave, so the wave is able to propagate in the plasma (see figure below). The collective electron motion is still able to cancel some of the electric field within one period of oscillation of the wave and establish a reflected radiation field as well.
- In the limit $\omega\gg \omega_p$, you are right that even the electrons hardly respond due to their inertia and establish only a weak radiation field. Thus the plasma behaves much like a vacuum with only very little reflection.
I hope this explains the sharp frequency dependence of the reflection and transmission coefficients. For a good description of radiation in a dielectric, I recommend reading this answer.
Plot of the dispersion relation for EM waves in a plasma: 
