Hi saltine!
Quote by saltine
I am confused because u is not a force but just acceleration.

Then it just goes onto the
acceleration side of F = ma …
you add it to the other (rotational) acceleration.
In Fig 1, there are two forces acting on the ball. The component of mg that is parallel to the rod is cancelled by the normall force from the rod, so the net force at the ball is only mg sinθ.

I didn't understand this at all.
The net force on the ball is mg sinθ
plus the force from the rod.
Nothing is cancelled (what is it to be cancelled with?) …
the reason why you leave out the mg cosθ is not because its cancelled, but simply because you're only looking at tangential components.
In Fig 5, the situation is similar except that u is now from an external force, F_{u} = Mu. Is this situation the same? Why is/isn't F_{u} part of the net force acting on the ball?

F
_{u} isn't "part of the net force acting on the ball" because isn't acting
on the ball at all …
you
must decide which body you're applying Netwon's second law to, and then only use the forces
on that body …
your book is using the ball as the body, so the only forces
on it are mg and the force from the rod …
any force on anything else is irrelevant.
Quote by saltine
Net Tangential Force = ma_{T} = mg sinθ  mu cosθ
Since a_{T} = α L,
m α L = mg sinθ  mu cosθ.
With small angle approximation:
[tex]mL\ddot{\theta} = mg\theta  mu[/tex]

This is much better … you've used the same method as your book, except that you've moved the mu cosθ from one side of the equation to the other, by using the accelerating frame.
(The tangential force is mg sinθ, and the tangential acceleration (times mass) is mαL + mu cosθ)
This
isn't necessary …
and it's risky, because it's so easy to get the ± sign wrong.
Your book just uses the stationary frame, and uses only tangential components, so that there's only
one component of force on the ball, and the acceleration is split into two parts.