I understand that your background is in physics, but (since I have neither time nor desire to write here a crash course in the relevant mathematics), I write my answer pretending that you are a reasonably advanced graduate student in math.
Let me first try to interpret your question: You start with the standard representation of the group $G=SO(3,1)_0$ (the connected component of identity in $SO(3,1)$, equivalently, the subgroup of $SO(3,1)$ preserving the future light cone) on the Lorentzian 4-space $R^{3,1}$, the 3-dimensional real vector space equipped with a nondegenerate symmetric bilinear form
$\langle, \rangle$ of signature $(3,1)$, with the associated quadratic form
$$
x^2 +y^2 + z^2 -w^2.
$$
The 2-fold nontrivial cover of the Lie group $G$ (the spinor group of $G$) is isomorphic to $SL(2, {\mathbb C})$ which has the natural (spinor) representation on ${\mathbb C}^2$. You would like to have an interpretation of vectors in ${\mathbb C}^2$ in terms of geometric objects associated with $R^{3,1}$.
a. First, I will do this in the case of the group $SO(2,1)_0$ since it is easier and illuminating. The natural representation of this group is on the Lorentzian 3-space $R^{2,1}$; the spinor group of $SO(2,1)_0$ is $SL(2, {\mathbb R})$ and its natural (spinor) representation is on ${\mathbb R}^2$. In order to interpret the corresponding spinors (vectors in ${\mathbb R}^2$), let me first note obstacles:
The dimension mismatch: 2 versus 3.
The action of $SL(2, {\mathbb R})$ on nonzero spinors is transitive (with stabilizers conjugate to the 1-parameter subgroup of strictly upper triangular matrices: Such matrices are called "unipotent"), while the action of $SO(2,1)_0$ on nonzero vectors in $R^{2,1}$ is non-transitive: There are three types of orbits, consisting of time-like, null and space-like vectors. Null-vectors will be most important for us, they are defined by the condition $\langle v, v \rangle=0$.
Any "natural" class of objects which one can define in $R^{2,1}$ will be acted upon by the group $SO(2,1)_0$ while we are interested in spinors, on which $SO(2,1)_0$ does not act.
While the space $C$ of future-like null-vectors is 2-dimensional (I regard the zero vector as a future null-vector for convenience), it does not have a natural structure of a vector space. Nevertheless, it does have a distinguished point (the origin) and a family of distinguished lines which are intersections of $C$ with affine 2-planes parallel to null-lines. The problem with these "lines" in $C$ is that some of them are only half-lines (the ones which pass through the origin); the rest are (connected) hyperbolas, so at least topologically they do look like lines.
Note, however, that:
The null-cone is 2-dimensional, homeomorphic to the 2-plane minus the origin. (Explicitly, you can define this homeomorphism by projecting $C$ to the coordinate plane in $R^{2,1}$ via the map $(x,y,z)\mapsto (x,y)$, where the quadratic form of the Lorentzian inner product is given by $x^2 + y^2 - z^2$.) This takes care of the dimension discrepancy.
The action of $SO(2,1)_0$ on the future null-cone $C$ minus the origin is transitive. The stabilizer of each nonzero null-vector is again 1-parameter (unipotent) subgroup of $SO(2,1)_0$. Under the covering map $SL(2, {\mathbb R})\to SO(2,1)_0$ the unipotent subgroups of the former map isomorphically to the unipotent subgroups of the latter. This is what we will exploit.
Now, I will let $S$ be the 2-fold cover of the punctured plane $N= C-\{{\mathbf 0}\}$. Informally, you can think of the elements of $S$ as null-vectors $v\in N$ each equipped with a "spin", a $\pm$ sign, which switches to the opposite sign as we rotate $v$ 360 degrees around the z-axis. In terms of polar coordinates we can think of the elements of $S$ as
$$
(r, \theta), r\ne 0, 0\le \theta< 4\pi.
$$
The reduction modulo $2\pi$ sends these to points in the punctured plane whose polar coordinates are
$$
(r, \theta), 0\le \theta< 2\pi.
$$
This passage to the 2-fold cover is the "unnatural step" which allows one to get spinors.
Now, the action of $SO(2,1)_0$ lifts to an action on $S$ but not of $SO(2,1)_0$ itself: The lift is the action of the spinor group $SL(2, {\mathbb R})$. One can verify (for instance, by observing that the action of $SL(2, {\mathbb R})$ on $S$ is transitive with point-stabilizers which are 1-parameter unipotent subgroups as required) that there is a diffeomorphism
$S\to {\mathbb R}^2 -\{{\mathbf 0}\}$ which conjugates the
action of $SL(2, {\mathbb R})$ on $S$ to the standard linear action of $SL(2, {\mathbb R})$ on ${\mathbb R}^2 -\{{\mathbf 0}\}$.
Under this diffeomorphism the "lines" which I mentioned above map to affine lines in $SL(2, {\mathbb R})$; each half-line lifts to the union of two half-lines in $S$ which map to a line (minus the origin) in ${\mathbb R}^2 -\{{\mathbf 0}\}$.
This gives you a reasonably geometric "Lorentzian" interpretation of nonzero spinors in ${\mathbb R}^2$ as elements of the surface $S$: These are nonzero future null-vectors $v\in N$ "equipped with a $\pm$ sign" to indicate which sheet of the 2-fold cover they lift to. The latter description is unsatisfactory as a mathematical description but should be OK as far as your intuition goes. The rigorous definition is in terms of covering spaces as I noted above. In order to get the zero spinor as well, one can simply say that we are using a 2-fold "branched cover" of $C$, which is ramified over the origin.
b. Now, to the Lorentzian 4-space $R^{3,1}$.The difficulties are somewhat similar. Again, note nontransitivity of the action of $G$ on the set of nonzero vectors in $R^{3,1}$. Inspired by (a), one can try to use the future null-cone $C\subset {\mathbb R}^{3,1}$. However, this results in the dimensional mismatch (the cone $C$ is 3-dimensional while the spinor space is real 4-dimensional). Also, while $G$ does act transitively on $C$ (minus the origin), the stabilizers are a bit larger than the ones in ${\mathbb C}^2$: The stabilizers of nonzero vectors in ${\mathbb C}^2$ are complex 1-parameter unipotent (real 2-dimensional), conjugate to the group of strictly upper triangular complex 2-by-2 matrices
$$
\left[\begin{array}{cc}
1&*\\
0&1\end{array}\right]
$$
while the stabilizers of nonzero null-vectors are 3-dimensional (in addition to 2-dimensional unipotent subgroups of $G$ which do lift to unipotent subgroups of $SL(2, {\mathbb C})$ we also have 1-parameter elliptic subgroups, isomorphic to $S^1$, which fix the null-vectors). Another problem is that $N= C- \{{\mathbf 0}\}$ is simply-connected, so taking its covering spaces would not be useful.
Nevertheless, what we can do is to take a future nonzero null-vector $v\in C$ and equip it with a half-plane $P$ which is tangent to the cone $C$ along the line spanned by $v$. Now, the stabilizer of each "flag" $(v,P)$ in $G$ is real 2-dimensional (a unipotent subgroup which lifts to a complex 1-dimensional unipotent subgroup of $SL(2, {\mathbb C})$ as required). I let $F$ denote the space of such "flags" $(v,P)$. It is not hard to check that this space is connected with the fundamental group isomorphic to ${\mathbb Z}_2$, which means that $F$ does have a connected 2-fold cover $S\to F$. One can also describe $F$ as the total space of the tangent bundle of the 2-sphere with the image of the zero section deleted. Now, we can play the same game as in (a): The elements of $S$ can be though of as flags $(v,F)$ equipped with a "spin", a $\pm$ sign which changes after we "spin" the half-plane $F$ around $v$ 360 degrees. One then verifies that the action of $G=SO(3,1)_0$ lifts to $S$ to an action of the spinor group $SL(2, {\mathbb C})$ on ${\mathbb C}^2$ minus the origin (again, by comparing the structure of point-stabilizers).
This is again a reasonably geometric description of (nonzero) spinors as elements of the 4-dimensional manifold $S$. The drawback of this description is that we do not see directly a complex structure on $S$ and the fact that the spinor group acts holomorphically; the linear structure is also very nontransparent. If this does not bother you, stop reading here; if it does bother you, proceed to the item (c).
c. I will now give a description of spinors which is derived from the Lorentzian geometry of ${\mathbb R}^{3,1}$, where the complex linear structure is transparent, but mathematics required to understand it gets harder.
Let's go back to the 3-dimensional null-cone $C\subset {\mathbb R}^{3,1}$. The space $\Sigma$ of future null-rays in $C$ is naturally diffeomorphic to the 2-sphere $S^2$; the action of $G$ on $C$ under the map of $\Sigma$ to $S^2$ becomes the conformal action of $PSL(2, {\mathbb C})$ on the Riemann sphere. The conformal structure on $\Sigma$ can be described as follows. For each nonzero null-vector $v\in N$, the restriction of the Lorentzian inner product to $T_vC$ (the tangent space of the cone $C$ at $v$, which is 3-dimensional) is degenerate positive semidefinite: The vector $v$ pairs to zero with each vector $w\in T_vC$. However, the projection $N\to \Sigma$ (sending the positive ray through $v\in C$ top a single point) divides out the line ${\mathbb R}v$ and, hence, $\langle, \rangle$ projects to a positive-definite inner product on the tangent plane to $\Sigma$ at the equivalence class $[v]$ of $v$. The action of $G$ on $\Sigma$ preserves the conformal class of the resulting Riemannian metric on $\Sigma$; the orientation is also preserved, hence, the action is conformal. (One can also describe the almost complex structure on $T\Sigma$ more directly but I will skip this.) The tangent bundle $T\Sigma$ is a complex one-dimensional vector bundle on $\Sigma$; algebraic geometers would call it the anticanonical bundle. It has degree $+2$, hence, there exists a "half-anticanonical line bundle" $L$ on $\Sigma$ so that the tensor square of $L$ is isomorphic to $T\Sigma$. The line bundle $L$ has degree $+1$; algebraic geometers call it the "hyperplane bundle" of ${\mathbb C}P^1\cong S^2$.
Remark. the total space $E$ of $L$ can be described as the 2-fold branched cover over $T\Sigma$ which is ramified over the image of the zero section of $T\Sigma$. Now, you may start to see a connection to Part (b). Fiberwise - this branched cover is nothing by the 2-fold branched cover over the complex plane ramifield at the origin. Now, you see a connection to Part (a).
The group $PSL(2, {\mathbb C})$ acts on $T\Sigma$ via (holomorphic) bundle automorphisms and this action lifts to an action of $SL(2, {\mathbb C})$ on $L$. It is easy to check that the space of holomorphic sections $\Gamma(L)$ of $L$ is complex-2-dimensional. The action of $SL(2, {\mathbb C})$ on the vector space $\Gamma(L)$ is manifestly complex linear, nontrivial. Hence, we obtain the spinor representation of $SL(2, {\mathbb C})$ on $\Gamma(L)\cong {\mathbb C}^2$.
A complex-analyst would describe sections of $L$ as "holomorphic half-vector fields" (or degree $-1/2$ holomorphic differentials)
$$
\omega=f(z)dz^{-1/2},
$$
where $f(z)$ as a holomorphic function. The strange degree $-1/2$ refers to the transformation law for such differentials: If $z=g(w)$ a conformal mapping then $g_*\omega$ is given by
$$
f(w) (g')^{-1/2} dw^{-1/2}.
$$
If $w=\frac{az+b}{cz+d}$ then
$$
f(w) (g')^{-1/2} dw^{-1/2}= f(w) (\frac{1}{(cw+d)^2})^{-1/2} dw^{-1/2} = f(w) (cw+d) dw^{-1/2}.
$$
Note that this expression is meaningless unless we specify the 2-by-2 matrix
$$
\left[\begin{array}{cc}
a&b\\
c&d\end{array}
\right]\in SL(2, {\mathbb C}).
$$
This is your spinor representation.
These holomorphic differentials of order $-1/2$ are (like it or not) your spinors. The linear structure is very transparent: In order to add these fellows, you just add the functional parts:
$$
f_1(z)dz^{-1/2} + f_2(z)dz^{-1/2}= (f_1(z)+ f_2(z))dz^{-1/2}.
$$
Linearity of the action of the spinor group is also clear (this action is just a change of variables in the differentials). The fact that the space of spinors is complex 2-dimensional might not be immediate but becomes clear once you think a bit about it. (The space is isomorphic to the space of holomorphic functions on the complex plane which have at worst simple pole at infinity, i.e. have the form $\alpha z + \beta$, $\alpha, \beta\in {\mathbb C}$.)
I am not sure of analytical importance of "holomorphic half-vector fields", but holomorphic half-order differentials
$$
\omega=f(z)dz^{1/2},
$$
do appear naturally in complex analysis when considering 2nd order linear holomorphic ODEs, that's how I first learned about them; see for instance (a bit dated but very clearly written):
N. S. Hawley and M. Schiffer,
Half-order differentials on Riemann surfaces,
Acta Math. Volume 115 (1966), 199--236.
Edit. See also Chapter 1 (The geometry of world-vectors and spin-vectors) in
Roger Penrose, Wolfgang Rindler, Spinors and space-time, Volume I, 1984.
