There are a number of mathematical imprecisions in your question and your answer. Some advice: you will be less confused if you take more care to avoid sloppy language.
First, the term spinor either refers to the fundamental representation of $SU(2)$ or one of the several spinor representations of the Lorentz group. This is an abuse of language, but not a bad one.
A particularly fussy point: What you've described in your first paragraph is a spinor field, i.e., a function on Minkowski space which takes values in the vector space of spinors.
Now to your main question, with maximal pedantry: Let $L$ denote the connected component of the identity of the Lorentz group $SO(3,1)$, aka the proper orthochronous subgroup. Projective representations of $L$ are representations of its universal cover, the spin group $Spin(3,1)$. This group has two different irreducible representations on complex vector spaces of dimension 2, conventionally known as the left- and right- handed Weyl representations. This is best understood as a consequence of some general representation theory machinery.
The finite-dimensional irreps of $Spin(3,1)$ on complex vector spaces are in one-to-one correspondence with the f.d. complex irreps of the complexification $\mathfrak{l}_{\mathbb{C}} = \mathfrak{spin}(3,1) \otimes \mathbb{C}$ of the Lie algebra $\mathfrak{spin}(3,1)$ of $Spin(3,1)$. This Lie algebra $\mathfrak{l}_{\mathbb{C}}$ is isomorphic to the complexification $\mathfrak{k} \otimes \mathbb{C}$ of the Lie algebra $\mathfrak{k} = \mathfrak{su}(2) \oplus \mathfrak{su}(2)$. Here $\mathfrak{su}(2)$ is the Lie algebra of the real group $SU(2)$; it's a real vector space with a bracket.
I'm being a bit fussy about the fact that $\mathfrak{su}(2)$ is a real vector space, because I want to make the following point: If someone gives you generators $J_i$ ($i=1,2,3$) for a representation of $\mathfrak{su}(2)$, you can construct a representation of the compact group $SU(2)$ by taking real linear combinations and exponentiating. But if they give you two sets of generators $A_i$ and $B_i$, then you by taking certain linear combinations with complex coefficients and exponentiating, you get a representation of $Spin(3,1)$, aka, a projective representation of $L$. If memory serves, the 6 generators are $A_i + B_i$ (rotations) and $-i(A_i - B_i)$ (boosts). See Weinberg Volume I, Ch 5.6 for details.
The upshot of all this is that complex projective irreps of $L$ are labelled by pairs of half-integers $(a,b) \in \frac{1}{2}\mathbb{Z} \times \frac{1}{2}\mathbb{Z}$. The compex dimension of the representation labelled by $a$,$b$ is $(2a + 1)(2b+1)$.
The left-handed Weyl-representation is $(1/2,0)$. The right-handed Weyl representation is $(0,1/2)$. The Dirac representation is $(1/2,0)\oplus(0,1/2)$. The defining vector representation of $L$ is $(1/2,1/2)$.
The Dirac representation is on a complex vector space, but it has a subrepresentation which is real, the Majorana representation. The Majorana representation is a real irrep, but in 4d it's not a subrepresentation of either of the Weyl representations.
This whole story generalizes beautifully to higher and lower dimensions. See Appendix B of Vol 2 of Polchinski.
Figuring out how to extend these representations to full Lorentz group (by adding parity and time reversal) is left as an exercise for the reader. One caution however: parity reversal will interchange the Weyl representations.
Sorry for the long rant, but it raises my hackles when people use notation that implies that some vector spaces are spheres. (If it's any consolation, I know mathematicians who get very excited about the difference between a representation $\rho : G \to Aut(V)$ and the "module" $V$ on which the group acts.)
Some comments:
First, let me describe the construction of the spinor bundle in a couple sentences. Given the smooth oriented Lorentzian manifold $(M,g)$, we have the tangent bundle $TM\to M$. We also have the $\mathrm{SO}(n,1)$ principal frame bundle, $FM\to M$. Now, $TM$ is the bundle associated to $FM$ via the natural representation of orthonormal frames as elements of $\mathrm{SO}(n,1)$. Since $\mathrm{SO}(n,1)$ has the double cover $\mathrm{Spin}(n,1)$, we suppose the existence of a double cover bundle of $FM$, $\mathrm{Spin}(n,1)\to P\to M$. We then take a vector space $\Delta$ on which there is a representation of $\mathrm{Spin}(n,1)$, this is where the Clifford algebra comes in. The spinor bundle is finally $S=P\times_{\mathrm{Spin}(n,1)}\Delta$. So the spinor bundle is a vector bundle associated to the double cover of the frame bundle.
The spin connection is indeed entirely dependent on the metric. The additional structure you added is the spin bundle (which need not be unique). The spin connection is in a sense a lift of the Levi-Civita connection from the tangent bundle.
The vielbeins $e^a{}_\mu$ are in a sense mixed frame-vector field objects. They can be thought of as representing the $TM\leftrightarrow FM$ duality from above.
The fibers of the spinor bundle are elements in the $\Delta$ from above, roughly speaking. They are spinors in the representation theory sense. The connection connects them in the same sense as for any Ehresmann connection.
Best Answer
A rotation of a spinor $\psi$ (looks like a complex 2-vector) by an angle $\phi$ around the unit axis $\hat n$ is but $$ \psi \mapsto e^{i {\phi\over 2} \left(\hat{n} \cdot \vec{\sigma}\right)} \psi= \left (I\cos {\phi\over 2} + i (\hat{n} \cdot \vec{\sigma}) \sin {\phi\over 2}\right ) \psi , $$ where $\vec \sigma$ are the three Pauli matrices, twice the generators of rotations in the doublet representation.
You can see how a 2π rotation amounts to flipping its sign, and twice that amounts to the identity.