In the traditional sense, acceleration means the rate at which velocity is changing. Visually, we tend to associate this with the magnitude of velocity.
If the centripetal acceleration of the moon is $2.8 \times 10^{-3} ~\mathrm {m/s^2}$, where the magnitude of velocity remains constant, what does this value mean? How can this value be interpreted intuitively, just as "regular" acceleration is so intuitive?
what does this value mean?
This value means for maintaining of moon on the circular path, we need a force by amount of $\textrm{mass of the moon}\times\textrm{that value}$
How can this value be interpreted intuitively?
According to first law of Newton, a moving body will move on a straight line uniformly unless a force is exerted on it.
As you can see in the figure, a force must exist so that pulls the moon towards the circle (otherwise, the moon won't move on the circle). The centripetal acceleration is this force divided by the mass of the moon.
Acceleration is defined as the time rate of change of velocity - be it in its direction or magnitude or both.
Velocity $\mathbf v$ in polar coordinates is given by:
$$\mathbf v= \dot {\mathbf r} = \dot r\mathbf{e_r} + r~\dot \theta\mathbf{e_{\theta}}\;.\tag 1$$
And then acceleration $\bf a$:
$$\mathbf a = \dot{ \mathbf v} = \left[\ddot r - r\dot \theta ^2\right]\mathbf{e_r} + \left[r\ddot \theta + 2~\dot r\dot \theta\right]\mathbf{e_\theta}\;.\tag 2$$
This is the crux of the kinematics involved in terms of polar coordinates.
Now, in circular motion, $\dot r= 0$ which leads $(2)$ to
$$\mathbf a = \dot{ \mathbf v} = - r\dot \theta ^2\mathbf{e_r} +r\ddot \theta \mathbf{e_\theta}\;.\tag {2-a}$$
The radial acceleration is the centripetal acceleration.
More precisely, in circular motion
$$\mathbf v= r\dot \theta \mathbf{e_\theta}$$
Therefore, the acceleration, is given by
$$\mathbf{a} = \frac{\mathrm d\mathbf v}{\mathrm dt}= r\frac{\mathrm d^2 \theta}{\mathrm dt^2}\mathbf{e_\theta} + \underbrace{r\frac{\mathrm d\theta}{\mathrm dt} \left(\frac{\mathrm d\mathbf {e_\theta}}{\mathrm dt}\right)}_\textrm{radial acceleration}\;.$$
So,
\begin{align}\mathbf{a_r} \equiv \textrm{centripetal acceleration} &= r\frac{\mathrm d\theta}{\mathrm dt} \left(\frac{\mathrm d\mathbf {e_\theta}}{\mathrm dt}\right)\\ &= v~(-\omega \mathbf{e_r})\;.\end{align}
Centripetal acceleration measures the time rate change of direction of the velocity vector i.e., $\mathbf{e_\theta}\;.$
Take a rope with a small weight on one end and spin around on a point such that the rope is taut. You will feel force on your arm 'pulling' on it towards the weight. That force exists because that's what you are exerting on the object to prevent constantly change it's direction. The value you refer to is the acceleration portion of that force from $F=ma$. If you attached a small spring scale to the rope, you could measure that force. One of my professors gave me a hard time (rightly, I suppose) about thinking about this from the perspective of the accelerated frame but it is much more intuitive to me.
