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.)
SU(2) representations
For representations of $SU(2)$, we know:
$$(2m+1) \otimes(2n+1) = (2(m+n)+1)\oplus (2(m+n-1)+1)\oplus...\oplus(2(m-n)+1)$$
We get this simply from angular momenta addition. For example:
$$2\otimes2 = 1 \oplus 3$$
Where the numbers label the dimension of (target space of) the representation (or the components of the representation).
This says the space that is invariant under the $2 \otimes 2$ representation, let's call is $S_2$ ( the target space of that representation), is isomorphic to the direct sum of 2 other spaces $S_1 \oplus S_3$, such that $S_1$ is invariant under the $1$ representation, and $S_3$ is invariant under $3$. In an explicit derivation for angular momenta addition in Hilbert space, $1$ would correspond to the 1-dimensional subspace of spin-0 states, $3$ corresponds to the 3-dimensional subspace of spin-1 states. The 1 spin-0 and 3 spin-1 spin eigenstates span the space, hence indeed the direct sum decomposition above holds.
Lorentz group representations
Lie algebra of the lorentz group is generated by some operators $L_i$, a useful decomposition is $L_i= N_i + N_i^{\dagger}$, where $N_i$, $N_i^{\dagger}$ are elements in an $SU(2)$ representation. Since $N_i$ and $N_i^{\dagger}$ commute, we can obtain the eigenspace of $L_i$ through angular momenta addition. Hence we can view the spaces spanned by the eigenvalues of each of the $N_k$ and $N_k^{\dagger}$ (for an arbitrary choice of k) as a Hilbert space, then $L_i$ is an operator in the tensor product of these Hilbert spaces. Namely, $L_i$ act the linear combinations of tensor product states $|... \rangle | ...\rangle$. This means we can view the representation $L_i$ as a tensor product of the 2 $SU(2)$ representations $N_i$ and $N_i^{\dagger}$. Hence the precise meaning of the notation $(1,2)$ is:
$$(a,b) \equiv a \otimes b$$
From here we can use the above result for tensor products of $SU(2)$ representations. For example:
$$(2,1)\otimes(2,1) = (2 \otimes 1)\otimes (2 \otimes 1) = (2 \otimes (1\otimes 2) \otimes 1) = (2 \otimes 2 \otimes 1)
\\=((1 \oplus 3)\otimes 1)= (1 \otimes 1) \oplus (3 \otimes 1) = (1,1) \oplus (3,1) $$
Where we've use the various properties of the tensor and direct products (associativity, distributivity). In the third equality we reduced the tensor product using the formula at the very top. Repeating the procedure for:
$$(2,1)\otimes(1,2)\otimes(2,2)$$
We clearly see $(1,1)$ is one of the terms in the direct sum decomposition.
Best Answer
Now when we pass to the full Lorentz group, somehow the projective representations disappear and become genuine representations.
I don't think this is true. Some but not all of the spinor representations of the proper orthochronous Lorentz group extend to representations of the full Lorentz group; you just add parity reversal and time reversal. But the new representations are still projective.