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.
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.
Best Answer
You can (and MUST, for spinors) indeed have both the spin connection and the gauge connection inside your covariant derivative. For example, for Dirac spinors with EM, you have:
$$D_\mu = \partial_\mu + 1/2 (\omega_{\alpha \beta})_\mu \sigma^{\alpha \beta} -ieA_\mu$$
It is more complicated than this and more mathematical involved than it is physically relevant, tell me in the comments if I need to edit this to make it more precise.
I will work with Euclidean signature just for convenience, the same is true for the full Lorentz group. The Lorentz group, here $SO(4)$, is used to build what is called an Orthonormal Frame Bundle. The fields are built using Associated Bundles. The covariant derivative lives in such a bundle. The connection always acts on representations of your gauge group. For bosonic fields you can use representations of $SO(4)$, so the connection is of vector-type, the usual Levi-Civita connection. For fermionic fields you can't use $SO(4)$ representations but you need to go to representations of its double cover. This gives you what is called a spin connection.
To build such a structure you will need also a Spin Structure.
It's important to note that you can think of the spin connection as the fundamental connection, remembering that on bosonic fields the representations of the $SO(4)$ group are equivalent to the ones of its double cover. In other words, you can, if you want to, express the Levi-Civita connection with tetrad instead of using the metric, because those gives you equivalent structures in vector-type representations