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.)
Your notion that a "spinor comes from an element in the Clifford algebra" seems misguided. (Dirac) Spinors are elements of $\mathbb C^4,$ as you say. $\rho$ doesn't turn Clifford algebra elements into (Dirac) spinors. It turns Clifford algebra elements into linear operators on Dirac spinors, also as you say ($\rho:\mathrm{Cl}(4)\to\mathrm{End}(\mathbb C^4)$). There's no meaningful way to get "a" unique spinor out of a Clifford algebra element.
In any case, the most important physical object when discussing spinors is the Lorentz group. We demand that physical quantities carry representations of the Lorentz group (projective representations, if quantum), because we demand that we can ("easily") predict the properties of an object after rotating or boosting it. Representations of the Clifford algebra are just a convenient proxy for (projective) representations of the Lorentz group: a copy of (the universal cover of) the Lorentz group sits inside $\mathrm{Cl}(4).$ This is strictly $$\mathrm{Spin}_\mathbb C(1,3)=\{v_1\ldots v_{2k}\mid v_i\in\mathbb C^4,|v_i|=1\}.$$ (Note: this is smaller than your $\mathrm{Cl}^0(4)$, since in addition to only taking the even elements I also impose a condition of unit norm. As $\rho$ is linear, this doesn't matter too much.) Yes, it may be true that, mathematically, the unrestricted representation $\rho:\mathrm{Cl}(4)\to\mathrm{End}(\mathbb C^4)$ does not split into two smaller representations. But this $\rho$ is not the physically relevant Dirac representation; not every element of $\mathrm{Cl}(4)$ represents a Lorentz transform. The Dirac representation is the restriction $\rho|_{\mathrm{Spin}_\mathbb C(1,3)},$ and this one does split.
That is, if you choose the right basis on $\mathbb C^4,$ you have $\rho|_{\mathrm{Spin}_\mathbb C(1,3)}(\Lambda)=\rho_L(\Lambda)\oplus\rho_R(\Lambda)$ for two distinct representations $\rho_L,\rho_R:\mathrm{Spin}_\mathbb C(1,3)\to\mathrm{End}(\mathbb C^2),$ the Weyl representations. I.e. there is more to the decomposition of a Dirac spinor into two Weyl spinors than just "2 + 2 = 4" in terms of dimensions. Specifically, a Dirac spinor rotates under Lorentz transforms as two independent objects of 2 dimensions each; the components of the two Weyl spinors don't get mixed up. (Compare to a 4-vector, which cannot be decomposed into 4 scalars, because Lorentz transforms mix them all up.)
Best Answer
For what it's worth, the defining/fundamental/spinor representation $$\bf{2}~\cong~\mathbb{C}^2~\cong~\mathbb{H}$$ of the Lie group $$Spin(3)~\cong~SU(2)~\cong~ U(1,\mathbb{H})$$ is a quaternionic/pseoudoreal representation, i.e. it is not a real representation.