Your second acceleration formula doesn't actually give the correct instantaneous acceleration - rather, an approximation of it.
Consider a more extreme case:
After the mass leaves point
, it will again return to
(after
seconds have passed).
The vector for change in velocity here is, of course, the zero vector (with magnitude zero).
This implies the acceleration is zero.
Now this is completely true (depending on what is meant by acceleration), but it's not the answer we want.
Why? Because this method only considers two points - the start and the end. The path of the mass was never taken into account.
Between
and
, the mass may as well have gone to the moon and back - as long as its initial and final velocities didn't change, we'd have no way of telling.
That's not to say that the second method is useless. With a bit of basic calculus, you can add a restriction of (uniform) circular motion and derive the centripetal force formula from it.