Rotation of a Spinor

The Pauli and Dirac matrices are basis vectors of Clifford algebras of 3d Euclidean space and 3+1d Minkowski space respectively. If you want to understand spinors, you'll probably need to understand Clifford algebras.

In Clifford algebras, reflections through the origin are represented by unit vectors (think of them as the surface normals of mirrors). The algebraic product composes reflections. Vectors can be written as weighted sums of basis vectors just as in the underlying vector space. In the Pauli/Dirac matrix representation, the Pauli/Dirac matrices are the basis vectors ($hat t$), $hat x$, $hat y$, $hat z$.

Any rotation can be written as a product of an even number of reflections. In 3d Euclidean space the Clifford products of even numbers of unit vectors live in a subspace of the algebra that's isomorphic to the unit quaternions. In 3+1d Minkowski space the subspace is isomorphic to the unit biquaternions.

To reflect a vector in a mirror, you multiply it on both sides by the Clifford representation of the surface normal (and possibly a factor of $-1$). You can convince yourself from the reflection interpretation of the algebra that this makes sense. To rotate a vector, you conjugate it by the appropriate even product, with the inverse being the same product in reverse order.

Spinors transform by multiplication by the same representations of reflections/rotations, but on only one side, not both.

I think a general geometric understanding of spinors is an open problem. However, at least in low dimensions (probably including 3+1), it's possible to think of the Clifford representation of a spinor as itself a rotation, from a "canonical spinor orientation" to the actual orientation. Rotating a spinor therefore means composing its representation with another rotation.

The essential reason that it takes a 720° rotation to get back to the original orientation is that reflection through two mirrors an angle $θ$ apart rotates an object by $2θ$. When you rotate a mirror through 180°, the plane of the mirror returns to its original position, but the surface normal points in the opposite direction, and the representation of the rotation as a product of vectors has therefore picked up a factor of $-1$.

Since you're asking about intuition my answer will be in that direction. You shouldn't mentally associate rotations with vectors on $mathbb{R}^3$. Instead, you should associate them with the group $SO(3)$. Moreover, you should think of $SO(3)$ as an abstract group, not as a set of $3times 3$ matrices. The set of 3x3 matrices is rather merely a representation of the group on 3 dimensional real space. This happens to be useful in classical physics where directions in space have 3 real degrees of freedom.

But apparently there are things in this world that have degrees of freedom that you may not be used to in the classical world, but are nonetheless very physical; such as the spin degree of freedom of an electron. These are described by a spinor (an element of $mathbb{C}^2$) and not a vector in $mathbb{R}^3$.

Now in physics we love linearisation and taylor expansions; they give you linear things and these are easy to work with. So instead of talking about translations and rotations we like to think about their first derivatives in a taylor expansion; namely linear velocities and angular velocities (which are proportional to the momenta and angular momenta). You may recognise angular velocities as (pseudo)vectors $omega_i$, but it's in fact equivalent and sometimes more convenient to express angular velocities as 3x3 antisymmetric matrices $W_{ij} = epsilon_{ijk}omega_k$. You may know what happens then when you exponentiate such an antisymmetric matrix: you get an $SO(3)$ matrix that makes up a finite rotation. More precisely in classical mechanics, you would write (with the poisson bracket and $L_z$ the angular momentum in the z direction): $R_phi = e^{{phi L_z, cdot}}$

In Quantum Mechanics, for spin degrees of freedom, you know the angular momentum operator in the z direction is $mathbf{frac{1}{2}}sigma_z$. Notice the half. The 2x2 complex rotation matrix by angle $phi$ around the z-axis is then just $R_phi = e^{iphifrac{1}{2}sigma_z}$. That half is what eventually leads to the half in the half angle argument and the $720^circ$ trick. Physically, this is just what happens to these $mathbb{C}^2$ degrees of freedom under rotation.

Personally, I don't really find it very surprising. The degrees of freedom of the spin of an electron just happen to transform someway under rotations. The angular velocity three degrees of freedom represented as a vector transform funnily under reflections, but there's nothing deep about it. The mass or temperature of an object for example don't transform at all under rotations, not surprising. The electron spin degrees of freedom transform funnily under rotations, why should that then be surprising.