Consider a scalar as having only magnitude (unit).
Consider a vector as having a magnitude and direction (unit up or unit down).
Now consider a spinor as having only direction (up and down).
So naturally, if a spinor only has direction, when you "rotate" anything it will change orientations, flip from up to down, the only rotation available to it.
Hence her use of "zero" not "null" vector as a zero vector has zero magnitude but still has direction while a null vector would have neither direction or magnitude (as I understand it).
Thus a spinor is "elementary" given that it embodies only direction, think of the "spin" of the electron or photon, while Euclidean space embodies both direction and magnitude.
I will answer your questions first defining the "classical" complex spinor representation of $Cl(\mathbb R^{2m})$, this is quite natural/geometrical being defined only in terms of wedging and contractions, it is related to complex manifolds and more generally to almost Hermitian manifolds. Then I will prove that this is isomorphic to the representation defined by Bär. This will answer question 2, the answer to question 1 will come up from the proof of the isomorphism.
I usually think of elements of the Clifford algebra as endomorphisms of the complex spinor bundle $S= \oplus_k \Lambda^{k,0}\mathbb R^{2m} $ aka the "complex spinor representation", the latter splits in positive and negative Weyl spinors as $S^+ = \oplus_{\text{k even}} \Lambda^{k,0}\mathbb R^{2m}$ and $S^- = \oplus_{\text{k odd}} \Lambda^{k,0}\mathbb R^{2m}$ the action of $Cl(\mathbb R^{2n})\otimes \mathbb C$ is defined by setting for $\alpha \in \Lambda^{k,0}\mathbb R^{2m}$
$$z_j \cdot \alpha = \sqrt{2} \ z_j \wedge \alpha$$
$$\bar z_j \cdot \alpha = -\sqrt{2} \ \alpha(z^j,\cdot)$$
(Recall that $\Lambda^{k,0} \mathbb R^{2m}$ is the complex vector space generated by $z_{i_1}\wedge \dots \wedge z_{i_k}$ for any subset $\{i_1,\dots, i_k\} \subset \{1,\dots, m\}$ of cardinality $k$.)
This is one of the most geometric ways of thinking about spinors because it ties them to complex geometry and shows that in that setting they are a natural object to consider.
Bär is defining a representation in a different way, instead of taking $S^+$ and $ S^-$ as above he considers his $\Sigma^\pm$ and defines the action by multiplication in the algebra.
One possible reason for doing so is that Bär's definition can be given without going through the definition of almost complex structures and forms of type $(p,q)$.
In any case, there is only one irreducible representation of $Cl(\mathbb R^{2n})$ up to isomorphism indeed there is an isomorphism of representations $F: S^\pm\simeq \Sigma^\pm$ defined as follows:
$$F(z_{i_1}\wedge \dots \wedge z_{i_k}) = 2^{-\frac k 2} z(i_1,\dots, i_k)$$
to see that $F$ interwines the action of the Clifford algebra we look at the behaviour under multiplication by $z_j$ and $\overline{z}_j$ (because these elements generate the complexified Clifford algebra $Cl(\mathbb R^{2n})\otimes \mathbb C$), we find that
$$\small{z_j \cdot (z_{i_1}\wedge \dots \wedge z_{i_k}) = 2^{1/2}z_j \wedge z_{i_1}\wedge \dots \wedge z_{i_k}\overset{F}{\mapsto }
2^{-(k+1-1)/2}z(j, i_1,\dots, i_k) = {z}_j \cdot F(z_{i_1}\wedge \dots \wedge z_{i_k})}$$
Similarly, we consider the case $j= i_1 $ for simplicity, the other cases are analogous, then
$$\small{\overline z_{i_1} \cdot (z_{i_1}\wedge \dots \wedge z_{i_k}) = - 2^{1/2} \iota_{z^{i_1}}(z_{i_1}\wedge \dots \wedge z_{i_k})\overset{F}{\mapsto }
-2^{-(k-2)/2}z(i_2,\dots, i_k) = \overline{z}_{i_1} \cdot F(z_{i_1}\wedge \dots \wedge z_{i_k})}$$
Where we used that
$$\overline{z}_{i_1}\cdot z(i_1,\dots, i_k) = (\bar z_{i_1}\cdot z_{i_1})z(i_1,\dots, i_k) = -2 z(i_1,\dots, i_k)$$
which relies on the followiing (painful to derive) identity in $Cl(\mathbb R^{2n})\otimes \mathbb C$:
$$(\bar z_{j}\cdot z_{j})\cdot \bar z_j = -\frac 1 2(1 + i e_j \cdot e_{j+1})\cdot\bar z_j = -2 \bar z_j.$$
(this is why the term $\bar z_1\dots \bar z_m$ appears in the definition of $z(i_1,\dots, i_k)$, so that we can use this identity)
This showes that the representation $S^\pm$ and $\Sigma^\pm$ are isomorphic.
A good reference for this is Section 3.4 of Friedrich.Dirac Operators in Riemmanian Geometry
Best Answer
There's a (hopefully) much less confusing way of viewing this, through the chirality operator (also defined in Hamilton's book), and denoted by $\Gamma$. It acts through the spinor representation on the space of Dirac spinors $\Sigma$, and its eigenvalues are $\pm 1$. From this, a direct sum decomposition is obtained $\Sigma = \Sigma_+\oplus\Sigma_-$. So every spinor $\sigma\in\Sigma$ can be written as $\sigma = \sigma_++\sigma_-$.
With respect to this decomposition, one has $\mathbb{C}l_0(n)\cong\text{End}(\Sigma_+)\oplus\text{End}(\Sigma_-)$, whereas $\mathbb{C}l_1(n)\cong\text{Hom}(\Sigma_+,\Sigma_-)\oplus\text{Hom}(\Sigma_-,\Sigma_+)$. In other words, even elements of the Clifford algebra preserve the parity (a.k.a. chirality) of the spinor, whereas odd elements reverse the parity.
The statement "... Dirac spinor that comes from an odd number of elements in the Clifford algebra..." does not make sense. Dirac spinors are elements of $\Sigma$, and they are acted upon by the Clifford algebra.