From here, how do I define the "resonance"?
At resonance, the energy flow from the driving source is unidirectional, i.e., the system absorbs power over the entire cycle.
When $\Omega = \omega_0$, we have
$$\phi(t) = \frac{A}{2\beta \omega_0}\sin\omega_0 t$$
thus
$$\dot \phi(t) = \frac{A}{2\beta}\cos\omega_0 t$$
The power $P$ per unit mass delivered by the driving force is then
$$\frac{P}{m} = j(t) \cdot \dot \phi(t) = \frac{A^2}{2\beta}\cos^2\omega_0 t = \frac{A^2}{4\beta}\left[1 + \cos 2\omega_0 t \right] \ge 0$$
When $\Omega \ne \omega_0$ the power will be negative over a part of the cycle when the system does work on the source.
What you've labelled as $\omega_r$ is the damped resonance frequency or resonance peak frequency.
Unqualified, the term resonance frequency usually refers to $\omega_0$, the undamped resonance frequency or undamped natural frequency.
This is just a footnote to Name's answer (which you should accept because it's correct) to give a slightly more intuition based argument.
If the driving force changes slowly compared to the natural frequency of the system then the system can move fast enough to stay in phase with the driving force. So most of the time the system is already moving in the direction the driving force is pushing it, and the force will accelerate the motion so the resulting amplitude of the oscillation will be big.
If the frequency of the driving force is a lot higher than the natural frequency of the system then the system cannot move fast enough to stay in phase with the driving force. This means some of the time the driving force is acting opposite to direction the system is moving, so it's slowing the motion not accelerating it. This means the amplitude of the motion will be reduced.
Best Answer
The oscillator frequency $\omega$ says nothing about the actual oscillator phase. Let us suppose that your oscillator oscillates freely like this: $$x(t) = A_0\cdot\cos(\omega t + \phi_0),\; t<0.$$ At $t=0$ it has a phase $\phi_0$. Depending on its value the oscillator can be moving forward or backward with some velocity. If you switch your external force on at $t=0$ and onwards, say, to push your particle in a positive direction, then, depending on the particle phase, the force will accelerate or decelerate the particle.
Generally you write down the external force in the same way: $$F_\text{ext}(t)=F_0\cdot \cos(\omega t +\Phi_0).$$ This expression stays in the driven oscillator equation, namely, in the right-hand side. The resonance happens always, i.e., the external force will supply energy to the particle, but this supplying can start immediately if the force direction and the particle velocity direction coincide. Otherwise the external force first slows down the particle and only then starts pumping its amplitude.
The particle velocity phase is shifted with respect to the particle coordinate $$v(t) = -A \sin(\omega t + \phi_0)=A\cos(\omega t + \phi_0 +\pi/2),\; t<0.\;$$ So, when $\Phi_0 =\phi_0 +\pi/2\;$ the force is in phase with velocity (not with coordinate) - you have not only the same direction for the velocity and for the force, but also coincidence of instants when both the velocity and the force become zero (no time intervals with their opposite signs).
EDIT: The permanent phase shift of $\pi/2$ in a resonant case with friction (as described in user17581 answer) is a self-established thing and its meaning is simple - the external force in the end compensates exactly the friction force; the latter being proportional to velocity which is shifted by $\pi/2$ with respect to the coordinate time-dependence (so the oscillator oscillates as if it were free, without losses).