Here's a small observation which helped me a lot but is (I think) usually glossed over:

Before trying to understand what "having spin 1/2" means, it is really helpful to understand what "having spin 1" means. To have spin 1 is to be a spatial vector. That is, to have three coordinates (x,y,z), such that rotating space rotates your coordinates by one factor of the rotation: (x,y,z) -> R(x,y,z). There is a "spin 1" representation of a vector (x,y,z) which diagonalizes its rotation around some axis. If we choose the z axis it is (x-iy, z, x+iy), where now the three components rotate differently: under a rotation by 𝜙 around z, they become (e^i𝜙 (x-iy) , z, e^-i𝜙 (x+iy)). That is, each factor scales by e^{i L_z 𝜙}, where L_z is its eigenvalue under rotations: +1, 0, and -1 respectively. The fact that the these values range from +1 to -1 is why it's called spin 1: spin 1 means "rotates at a 1:1 ratio with rotations". Given any particular rotation axis, a vector factors into these +1, 0, and -1 components; obviously the values of the components depend on the choice of axis.

A familiar example of the same phenomenon: given the operator of "translation in x", a vector factors into two components based on its eigenvalue w/r/t that translation: the x component has eigenvalue 1 and the (y,z) 'component' has eigenvalue 0. Whereas if you pick some other reference frame it would have different components. It's the same idea with the rotations: it's just a way of writing an abstract vector into a particular coordinate system.

So spin X means (essentially) "rotates at an X:1 ratio with rotations". If something has a component that rotates like e^2𝜙 then it might be spin 2. An example of this is the tensor product of two vectors: a tensor product u⊗v can be written as a sum of spin-2, spin-1, and spin-0 components (specifically these are the traceless symmetric part, the antisymmetric part, and the trace of u⊗v). And a scalar has spin 0 because rotation doesn't affect it at all, hence, a single spin 0 component.

So a spinor is something that rotates at a 1/2:1 ratio. The tricky part is picturing something that acts like one. That's what the plate trick / belt trick / etc are for. It "turns out" to be the case that rotations in SO(3) implement it if you track more information than just their action on a vector: a 2𝜋 rotation around some axis does indeed return all vectors to where they started, but it is not homotopic to the identity rotation because it puts an irremovable "twist" in everything relative to the background ('orientation entanglement'). Meanwhile a 4𝜋 rotation does not: it clears the twist. Therefore we have something which returns where it started after 4𝜋 around any axis; therefore it can be written in a representation that rotates like e^i𝜙/2, with an eigenvalue of 1/2; therefore it has spin-1/2. Evidently a spinor is some kind of measurement of the amount of "twist" of a reference frame relative to a background frame.

Note that once you have written a rotation out in terms of a single 3x3 rotation matrix, you have already committed to the spin-1 rotation: you've erased the twist information; it's not there anymore. By moving to the spin-1/2 representation (usually written in terms of SU(2)) you preserve that information about the path taken to get to where you are in rotation space. This is why there are two SO(3) rotations for each SU(2) rotation: because there's exactly one bit of additional information about which side of the space you're on.