Is there a difference between "average acceleration" and centripetal acceleration?
Yes, in fact they're almost completely unrelated.
The average acceleration is defined as
$$\vec a_\text{avg} = \frac{\Delta\vec v}{\Delta t}$$
It is one quantity that partially describes the motion of a particle over an extended time. In other words, average acceleration encapsulates the fact that a particle started with some velocity at time A and ended with some velocity at time B, but completely ignores what the particle did between A and B. This is by design.
Centripetal acceleration, on the other hand, is an instantaneous quantity: it's the radial component of acceleration. (This requires that you have chosen some point to be the center of a polar coordinate system.) It partially describes the motion of a particle at one moment, not over an extended time.
The conventions of differential calculus are subtle and hard to explain,
and that seems to be the core of this conundrum. Note, 'velocity', a vector, is
defined at every point of the trajectory of a moving object, and is defined
as a limit of very-short-path-traveled divided by very-short-time-elapsed.
So, the magnitude of the velocity, the speed, and the direction of the velocity
(which, in circular motion, just takes on every value possible in two dimensions),
are both important in general, but in the specific case of a complete
circle, only magnitude has a non-trivial meaning. Expressing a formula
with "v^2" reinforces that only the magnitude is intended, because
there is no general way to 'square' a vector.
Acceleration, like velocity, is a vector defined at every point, by a similar
limit of very-small-velocity-change divided by very-short-time-elapsed,
and is inconsistent with using an AVERAGE of velocity (since it depends
on the change over very short times). Acceleration has a direction
just as velocity does, but (again, because this is a circle) it is
described as 'centripetal' meaning it takes the direction toward the
center of the circle.
It makes little sense to talk of average directed acceleration, because
the full circular path makes that average vanish (by symmetry) in uniform
circular motion: only the magnitude of the acceleration is constant
in this kind of motion. So, the "centripetal acceleration" means
only the magnitude of the instantaneous acceleration, without any
averaging.
So, one cannot make use of the given formula while averaging vectors over
a whole cycle of the circular motion. Differential calculus makes
the hints, and therefore the formula for acceleration, more understandable.
Best Answer
The centripetal acceleration is the instantaneous acceleration which is given by $$\vec{a}_{\rm ins}=-\frac{v^2}{r}\hat{r}$$ which has a magnitude of $9$ in your case, but is constantly changing in direction.
The average acceleration is defined by $$\vec{a}_{\rm ave}=\frac{\Delta \vec{v}}{\Delta t}$$ which has magnitude of $6$ in your case and is pointing to the right.
By definition $$\vec{a}_{\rm ins}(t)=\lim_{\Delta t \to 0}\frac{\vec{v}(t+\Delta t)-\vec{v}(t)}{\Delta t}=\lim_{\Delta t \to 0}\vec{a}_{\rm ave}$$ or equivalently $$\vec{a}_{\rm ave}=\frac{1}{\Delta t}\int_t^{t+\Delta t}\vec{a}_{\rm ins}dt$$
In your case, you can interpret it this way: Although the instantaneous centripetal acceleration has a constant magnitude of $9$, it is constantly changing in direction. Hence there are some cancellations when you find the vector sum in the integral above, which leads to an average acceleration of magnitude $6$ pointing to the right.