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.
Fiber bundles can have nontrivial topology, but for this question, we might as well work in a patch where it's trivial. Also, this answer uses classical fields to avoid possible technical issues when dealing with field operators (or distributions) on a Hilbert space.
For any analytic function $f(x)$, we have the identity
$$
f(x+c)=e^{c\cdot\partial}f(x)
\tag{1}
$$
where $\partial$ is the derivative with respect to $x$ and $c$ is independent of $x$. We might be tempted to write this as
$$
f(y)=e^{(y-x)\cdot\partial}f(x),
\tag{2}
$$
but that's not correct unless we interpret it like this:
$$
f(y)=\Big[e^{(y-x_0)\cdot\partial}f(x)\Big]_{x_0=x}.
\tag{3}
$$
Either way, it doesn't matter whether $f$ is a component of a gauge field (in flat or curved spacetime) or a component of a spinor field (in a gauge theory or not). In other words, it doesn't matter what the function $f$ represents. Equations (1) and (3) are identities that hold for any analytic function $f$. By the way, regarding the distinction between analytic and smooth, see https://en.wikipedia.org/wiki/Non-analytic_smooth_function.
In contrast, covariant derivatives are used to define new fields that preserve certain properties, which can facilitate the construction of gauge-invariant lagrangians. Consider a spinor field $\psi$ coupled to an abelian gauge field $A$. I'm using an abelian gauge field in this example so that we don't need to worry about path-ordering. Suppose we define a new field $\psi'$ by
$$
\psi'(x+c)=\psi(x+c)-\exp\left(i\int_x^{x+c} dy\cdot A(y)\right)\psi(x).
\tag{4}
$$
Under a gauge transform
\begin{gather}
\psi(x)\to e^{i\theta(x)}\psi(x)
\\
A_\mu(x)\to A_\mu(x)+\partial_\mu\theta(x),
\tag{5}
\end{gather}
the field $\psi'$ transforms as
$$
\psi'(x)\to e^{i\theta(x)}\psi'(x),
\tag{6}
$$
just like $\psi$ does. For infinitesimal $c$, equation (4) becomes
\begin{align}
\psi'(x+c)
&=\psi(x)+c\cdot\partial\psi(x)-ic\cdot A\psi(x) + O(c^2) \\
&=\psi(x)+c\cdot D\psi(x) + O(c^2)
\tag{7}
\end{align}
with $D_\mu=\partial_\mu-iA_\mu$. By the way, the finite version (4) is used in lattice gauge theory, where derivatives are replaced by finite differences.
I don't have a copy of Greiner's book on hand, but I'm guessing it's using the same symbol $\psi$ for both functions, writing $\psi$ instead of $\psi'$ on the left-hand side of (7). Yeah, that's careless, but it's also common: people often use the same symbol for different-but-related things. It's a compromise to avoid a proliferation of primes, tildes, hats, subscripts/superscripts, and other decorations when several different related quantities are involved. I don't know if that's what Greiner's book is doing here, but it's a plausible guess.
Best Answer
A gauge field for a particular group $G$ can be thought of as a connection, or a $G$ Lie algebra valued differential form. If we recall the Riemann curvature,
$$R(u,v)w = \left( \nabla_u \nabla_v - \nabla_v \nabla_u -\nabla_{[u,v]}\right)w$$
If $[u,v]=0$ the expression simplifies to the usual tensor in general relativity. Similarly, we may think of the field-strength of a gauge field as a curvature - it's essentially a commutator of covariant derivatives and attempts to quantify the affect of parallel transportation on tensorial objects. For a $U(1)$ field,
$$F=\mathrm{d}A $$
with no additional terms, because the analogue of the $\nabla_{[u,v]}$ term vanishes as $U(1)$ is abelian and all structure constants of the group vanish. The relation to the curvature tensor becomes even clearer as we express the field-strength in explicit index notation,
$$F=\partial_\mu A_\nu - \partial_\nu A_\mu$$
In gravitation, the gauge group is the group of diffeomorphisms $\mathrm{Diff}(M)$, infinitesimally these are vector fields which shift the coordinates; the binary operation of the group is the Lie bracket, and the metric changes by a Lie bracket, namely,
$$g_{ab}\to g_{ab}+\mathcal{L}_\xi g_{ab}$$
where $\xi$ is our vector field. The Lorentz group $SO(1,3)$ is a subgroup of the diffeomorphism group. In addition, the Killing vectors are those which produce no gauge perturbation of the metric, i.e.
$$\nabla_\mu X_\nu -\nabla_\nu X_\mu=0$$
These Killing vector commutators may form a Lie algebra of a Lie group $G$; the generators $T_a$ of a Lie group $G$ allow us to define the structure constants,
$$[T_a,T_b]=f^{c}_{ab}T_c$$
where $f$ are the structure constants, modulo some constants according to convention.