Here we define unit vector n in the polar coordinate.
[tex]\vec{n} = ( \sin\theta \cos\varphi, \, \sin\theta \sin\varphi, \, \cos\theta )[/tex]
Of course by 2pi rotation, this vector doesn't change,
[tex]\varphi \to \varphi + 2\pi, \qquad \vec{n} \to \vec{n}[/tex]
The n component of the spinor operator is
[tex]\hat{S}_n = \vec{n}\cdot \vec{S} = \frac{\hbar}{2} \left[ (\sin\theta \cos\varphi) \sigma_1 + (\sin\theta \sin\varphi) \sigma_2 + (\cos\theta) \sigma_3 \right][/tex]
where sigma is Pauli matrices. So,
[tex]\hat{S}_n = \frac{\hbar}{2} \left( \begin{array}{cc} \cos\theta & e^{-i\varphi} \sin\theta \\ e^{i\varphi} \sin\theta & -\cos\theta \end{array} \right)[/tex]
The eigenstate of this operator (which direction is "n" ) is
[tex]\alpha_n = \left( \begin{array}{c} \cos \frac{\theta}{2} e^{-i\varphi /2} \\ \sin\frac{\theta}{2} e^{i\varphi /2} \end{array} \right) \quad \hat{S}_n \alpha_n = \frac{\hbar}{2} \alpha_n[/tex]
If we rotate the direction of "n" by 2pi, the unit vector "n" doesn't change, as shown above.
[tex]\vec{n} \to \vec{n} \quad (\varphi \to \varphi + 2\pi)[/tex]
But its eigenstate change from +1 to -1 by 2pi rotation.
[tex]\alpha_n \to - \alpha_n \quad (\varphi \to \varphi + 2\pi)[/tex]