Short version : ${\omega_\kappa}^\mu$ is a matrix of (dual) vectors, and it is basically identical to the Christoffel symbol, via each such vector being
\begin{equation}
{\omega_\kappa}^\mu = ({\Gamma^\mu}_{\kappa 0}, {\Gamma^\mu}_{\kappa 1}, \ldots)
\end{equation}
In that paper in particular, they use the dual basis $\{ \omega^\nu \}$, so that
\begin{equation}
{\omega_\kappa}^\mu = {\Gamma^\mu}_{\kappa 0} \omega^0 + {\Gamma^\mu}_{\kappa 1} \omega^1 + \ldots
\end{equation}
Although please note, that basis may not be a coordinate basis, so it should be more exactly
\begin{equation}
{\omega_\kappa}^\mu = {\Gamma^\mu}_{\kappa 0} e^0_{\nu}\omega^\nu + {\Gamma^\mu}_{\kappa 1} e^1_{\nu} \omega^\nu + \ldots
\end{equation}
Long version : The covariant derivative can be seen as a map from the tangent bundle to the product of the tangent bundle with the exterior bundle of $1$-forms,
\begin{eqnarray}
\nabla : TM &\to& TM \otimes \Omega^1 M\\
X &\mapsto& \nabla X
\end{eqnarray}
That way, we can define the application of a vector to the covariant derivative, $\nabla_Y$X, as an application of that $1$-form to $\nabla_X$, ie $[\nabla X](Y) = \nabla_Y X$.
Under this form, we can in particular define, given a function $f$ and a vector field $v$,
\begin{eqnarray}
\nabla(fv) = df \otimes v + f \nabla v
\end{eqnarray}
with the application
\begin{eqnarray}
[\nabla(fv)](Y) = df(Y) \otimes v + f [\nabla v] (Y)
\end{eqnarray}
We can always decompose the vector on some frame field $\{ e_\mu \}$, so that
\begin{eqnarray}
v = v^\mu e_\mu
\end{eqnarray}
The components are then just a set of four scalar functions. We then have
\begin{eqnarray}
\nabla(v^\mu e_\mu) = dv^\mu \otimes e_\mu + v^\mu \nabla e_\mu
\end{eqnarray}
$\nabla e_\mu$ is once again a covariant derivative, so it will be equal to some object in $TM \otimes \Omega^1 M$. That object is the connection form. The vector part of this object can itself be decomposed in components
\begin{eqnarray}
\nabla e_\mu = \omega_\mu^\nu e_\nu
\end{eqnarray}
Each component here $\omega_\mu^\nu$ is a $1$-form. It can be decomposed further in a slightly more obvious vector/$1$-form decomposition by considering the dual basis $\theta^\mu$, which is the basis of $1$-forms such that $\theta^\mu(e_\nu) = \delta^\mu_\nu$ :
\begin{eqnarray}
\omega_\mu^\nu e_\nu &=& (\omega_\mu^\nu)_\sigma e_\nu \otimes \theta^\sigma
\end{eqnarray}
In that decomposition, this is basically the Christoffel symbol. Just apply a vector $Y = Y^\alpha e_\alpha$ and apply linearity and the Leibniz rule to it to see that :
\begin{eqnarray}
[\nabla(X^\mu e_\mu)](Y^\alpha e_\alpha) &=& dX^\mu(Y^\alpha e_\alpha) \otimes e_\mu + X^\mu [\nabla e_\mu](Y^\alpha e_\alpha)\\
&=& Y^\alpha \left[dX^\mu(e_\alpha) \otimes e_\mu + X^\mu (\omega_\mu^\nu)_\sigma e_\nu \otimes [\theta^\sigma](e_\alpha)\right]\\
&=& Y^\alpha \left[\partial_\alpha X^\mu e_\mu + X^\mu (\omega_\mu^\nu)_\sigma e_\nu \delta^\sigma_\alpha\right]\\
&=& Y^\alpha \left[\partial_\alpha X^\mu e_\mu + X^\mu (\omega_\mu^\nu)_\alpha e_\nu \right]
\end{eqnarray}
You will recognize the appropriate form of the covariant derivative, so that
\begin{eqnarray}
(\omega_\mu^\nu)_\alpha = {\Gamma^\nu}_{\mu\alpha}
\end{eqnarray}
The link between the two is then that for any two components $\mu, \nu$, the connection form is a $1$-form equal to
\begin{eqnarray}
\omega_\mu^\nu = {\Gamma^\nu}_{\mu\alpha} \theta^\alpha
\end{eqnarray}
