Difference tensor between two connections
OP's question is probably spurred by the fact that Ref. 1 forgets to mention that:
The other connection $\nabla:\Gamma(TM)\times \Gamma(TM)\to \Gamma(TM) $ with lower Christoffel symbols $$\Gamma_{ij,k}~:=~ \Gamma_{ij}^{\ell}~g_{\ell k}\tag{1}$$ is still assumed to be compatible with the metric $$\nabla g = 0\qquad\Leftrightarrow\qquad \partial_kg_{ij}=\Gamma_{ki,j}+\Gamma_{kj,i}.\tag{2}$$
The contorsion tensor $K~\in~\Gamma\left(T^{\ast}M\otimes \bigwedge^2(T^{\ast}M)\right)$ is defined$^1$ as the difference between the metric-compatible $\nabla$ and the Levi-Civita connection $\nabla^{LC}$, i.e. $$K_{i,jk}~:=~\Gamma_{ij,k}-\Gamma^{LC}_{ij,k} ~\stackrel{(2)}{=}~-K_{i,kj}.\tag{3} $$
The torsion tensor $T~\in~\Gamma\left(\bigwedge^2(T^{\ast}M)\otimes T^{\ast}M \right)$ is defined$^1$ as $$ T(X,Y)~:=~\nabla_XY-\nabla_YX-[X,Y] $$ $$\quad\Leftrightarrow\quad T_{ij,k}~:=~\Gamma_{ij,k}-\Gamma_{ji,k}~\stackrel{(3)}{=}~K_{i,jk}-K_{j,ik}.\tag{4} $$
One may show that the inverse relation to (4) is $$ K_{i,jk}~\stackrel{(4)}{=}~ \frac{1}{2}\left(T_{ij,k}-T_{jk,i}+T_{ki,j}\right),\tag{5}$$ and vice-versa.
Let us decompose the contorsion tensor $$K_{i,jk} ~=~ \frac{1}{2}(K^+_{i,jk}+K^-_{i,jk}), \tag{7}$$ in components $$ K^{\pm}_{i,jk}~:=~K_{i,jk}\pm K_{j,ik}, $$ $$ K^+_{i,jk}~=~T_{ki,j}+T_{kj,i}, \qquad K^-_{i,jk}~=~T_{ij,k}, \tag{8} $$ that are symmetric/antisymmetric wrt. the first two indices $i\leftrightarrow j$.
The geodesic equation reads $$0~=~\nabla^{LC}_{\dot{\gamma}}\dot{\gamma} \qquad\Leftrightarrow\qquad \ddot{\gamma}^{\ell}~=~-\Gamma^{LC,\ell}_{ij}\dot{\gamma}^i\dot{\gamma}^j$$ $$\qquad\Leftrightarrow\qquad -g_{k\ell}\ddot{\gamma}^{\ell}~=~\Gamma^{LC}_{ij,k}\dot{\gamma}^i\dot{\gamma}^j.\tag{9} $$
In contrast, the auto-parallel equation $$0~=~\nabla_{\dot{\gamma}}\dot{\gamma} \qquad\Leftrightarrow\qquad \ddot{\gamma}^{\ell}~=~-\Gamma^{\ell}_{ij}\dot{\gamma}^i\dot{\gamma}^j$$ $$\qquad\Leftrightarrow\qquad -g_{k\ell}\ddot{\gamma}^{\ell}~=~\Gamma_{ij,k}\dot{\gamma}^i\dot{\gamma}^j~=~\left(\Gamma^{LC}_{ij,k}+\frac{1}{2}K^+_{i,jk}\right)\dot{\gamma}^i\dot{\gamma}^j\tag{10} $$ can detect the symmetric part $K^+_{i,jk}$ of the contorsion.
Observation: $$T_{ij,k} \text{ is totally antisymmetric} \qquad\Leftrightarrow\qquad K_{i,jk} \text{ is totally antisymmetric} $$ $$\qquad\Leftrightarrow\qquad K^+_{i,jk}~=~0.\tag{11}$$ In the affirmative case, we have $T_{ij,k}=2K_{i,jk}$, and the geodesic & auto-parallel equations (9) & (10) are the same.
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$^1$ Pertinent applications of the musical isomorphism are implicitly implied from now on.