I think what's important to consider here is that the Levi-Civita connection is unique. Once you introduce torsion this is no longer true. Therefore, when you write $g_{ij;k}$ it isn't clear if you're considering the derivative with the Levi-Civita connection, or if you're considering a derivative operator with torsion. However, I think the answer to your question may be the following.
If you consider a derivative operator $\nabla_a$ such that
$$[\nabla_a, \nabla_b]f = -T^c_{\ ab} \nabla_c f,$$
where $T^c_{\ ab}$ is the torsion tensor, and for an arbitrary one-form $\omega_a$ you have
$$\nabla_a \omega_b = \partial_a \omega_b - C^c_{\ ab} \omega_c,$$
where $C^c_{\ ab}$ is the connection associated with our derivative operador $\nabla_a$; if you suppose $\nabla_a$ is metric compatible then you can show that
\begin{align}
C^c_{\ ab} &= \frac{1}{2} g^{cd} (\partial_a g_{bd} + \partial_b g_{ad} - \partial_d g_{ab}) + \frac{1}{2} (T^c_{\ ab} + T^{\ c}_{a \ \ b} + T^{\ c}_{b \ \ a})\\
&= \Gamma^c_{\ ab} + K^c_{\ ab},
\end{align}
where $\Gamma^c_{\ ab}$ is the Christoffel symbol and $K^c_{\ ab}$ is the contorsion tensor. In fact, from this expresion you can see that if $D_a$ is the derivative operator associated with the Levi-Civita connection, then
$$(D_a - \nabla_a) \omega_b = K^c_{\ ab} \omega_c.$$
Thus, the contorsion tensor is the connection between the derivative operator associated with the Levi-Civita connection and the derivative operator with torsion.
Edit: I'd also like to give a bit of insight into the last part of your question. If you consider the Einstein-Cartan action with an additional matter action
$$S = S_{\text{EC}} [g^{ab}, K^c_{\ ab}] + S_{\text{M}} [g^{ab}, K^c_{\ ab}, \phi],$$
where $\phi$ is an arbitrary matter field, and
$$S_{\text{EC}} = \frac{1}{16 \pi} \int \mathcal{R} \sqrt{-g} \text{d}^4 x,$$
where $\mathcal{R}$ is the Ricci scalar associated with the derivative operator $\nabla_a$; the resultant equations of motion are
\begin{align}
&G_{ab} + K^d_{\ (ab)} K^c_{\ cd} - K^d_{\ c(b} K^c_{\ a)d} - K^{d \ \ c}_{\ [c} K^e_{\ e]d} g_{ab} = 8 \pi \tau_{ab},\\
&g^{ab} K^d_{\ dc} + \delta^a_{\ c} K^{bd}_{\ \ \ d} - K^{ab}_{\ \ \ c} - K^{b \ \ a}_{\ \ c} = 16 \pi \sigma_c^{\ \ ab},\\
&\frac{\delta (\mathcal{L}_{\text{M}} \sqrt{-g})}{\delta \phi} = 0.
\end{align}
Note that $G_{ab}$ is the Einstein tensor associated with the Levi-Civita derivative operator $D_a$. The stress-energy tensor $\tau_{ab}$ is defined as
$$\tau_{ab} := -\frac{2}{\sqrt{-g}} \frac{\delta (\mathcal{L}_{\text{M}} \sqrt{-g})}{\delta g^{ab}}.$$
The spin density $\sigma_c^{\ ab}$ is defined as
$$\sigma_c^{\ \ ab} := -\frac{1}{\sqrt{-g}} \frac{\delta (\mathcal{L}_{\text{M}} \sqrt{-g})}{\delta K^c_{\ ab}}.$$
What's important is that the equation of motion for the contorsion tensor is algebraic, there are no derivatives involved. It can also be shown that you can write the contorsion tensor in terms of the spin density. This means that if the spin density is zero, then the contorison tensor is zero and viceversa. Outside the region of a given concentration of matter, there's no spin density, which in turn means the contorsion tensor vanishes, and you are left with $G_{ab} = 8 \pi \tau_{ab}$; which is usual GR.
In the case of the Schwarzschild metric as well as the Kerr metric, you're working with a vacuum solution, which implies $\sigma^c_{\ ab} = 0$, which in turn implies $K^c_{\ ab} = 0$, and therefore you obtain the same dynamics that usual GR describes in the black hole's exterior region.
