Question: I have an example that I don't understand, especially 2 expressions.

Given a spin in the state $|z + \rangle$, i.e., pointing up along the z-axis what are the
probabilities of measuring $\pm \hbar/2$ along $\hat{n}$?

I have these 2 expressions

$$|+\rangle\_n = cos(\theta/2)e^{-i\phi/2}|+\rangle +sin(\theta/2)e^{i\phi/2}|-\rangle$$
$$|-\rangle\_n = sin(\theta/2)e^{-i\phi/2}|+\rangle -cos(\theta/2)e^{i\phi/2}|-\rangle$$

I'm not sure to fully understand what that means.
Is $|+\rangle\_n$ only used for a particle in the spin $|z + \rangle$ and $|+\rangle\_n$ is a linear combination of $\hbar/2$ and $- \hbar/2$.
Hence, the probability to have $\hbar/2$ is $|cos(\theta/2)e^{-i\phi/2}(1,0)|^2$, knowing that $|+\rangle = (1,0)$

I'm not that convince about my explanation. This is not based on anything.

Answer: 
If there is a non rotating object (like a pole) in the center of the merry go round, this object will look rotating around itself, from the edge person perspective.

But all other non rotating objects (the rest of the world) will rotate around the pole in the center.

It is easy to see that for non rotating objects close to the person in the edge at some moment. Of course those object will be far away after half a cycle. So they are not rotating around the person.
I think the notion of "actual centre" depends on the frame in which you are considering your rotation motion.

For an observer outside of the rotating frame, the merry-go-round actually has a centre about which it rotates. But when we shift frames to the person on the mere-go-round, the centre in the rotating frame would be the point where all the fictitious forces which are experienced by a person in the rotating frame vanish (i.e the actual centre of rotation).

Thus the centre of the rotation is the same irrespective of who is viewing the motion, the difference is only how you describe that point.

I hope this helps.