I have found an apparent contradiction by using abstract index notation. I supose I am doing something wrong, but I cannot find the mistake. The question is the following:
Given two connetions $\check\nabla$ and $\nabla$ (not necessarily metric connections), let's define the transition tensor $Q^c_{ab}$ by $\check\nabla_a v^c -\nabla_av^c=Q^c_{ab}v^b$. Given that $(\check\nabla_a -\nabla_a)f=0$ for every scalar function, it can be easily proved that the transition tensor verifies $\check\nabla_a v_b -\nabla_av_b=-Q^c_{ab}v_c$ for dual fields.
I want now to relate the transition tensor with the difference of torsions $\check\Sigma -\Sigma$ for the connections $\check\nabla$ and $\nabla$ respectively. By definition, the torsion verifies:
$$\Sigma^c_{ab}\nabla_cf=\nabla_a\nabla_bf-\nabla_b\nabla_af.$$
Given that $\nabla_cf =\omega_c$ is a dual field, we can see:
$$ \check\Sigma^c_{ab}\omega_c-\Sigma^c_{ab}\omega_c=(\check\nabla_a -\nabla_a)\omega_b - (\check\nabla_b -\nabla_b)\omega_a=(-Q^c_{ab}+Q^c_{ba})\omega_c,$$ that is $ \check\Sigma^c_{ab}-\Sigma^c_{ab} = -Q^c_{ab}+Q^c_{ba} = -2Q^c_{[a,b]} $.
On the other hand, if I use the index free notation, the torsion is:
$$ \Sigma(\bf{u},\bf{v})=\nabla_{\bf{u}}\bf{v}-\nabla_{\bf{v}}\bf{u}-[\bf{u},\bf v],$$ therefore it is straightforward that
$$ (\check\Sigma - \Sigma)(\bf{u} , \bf{v}) = Q(\bf{u}, \bf{v})-Q(\bf v, \bf u).$$ Then $ \check\Sigma^c_{ab}-\Sigma^c_{ab} = Q^c_{ab} - Q^c_{ba} = 2Q^c_{[a,b]},$ has different sign from the above expression.
This sign is crucial to cancel terms in further calculations and the correct one is the first (negative). I cannot find the mistake. It could be the definition of $Q$ in the index free notation, which I guess is
$$ Q(\bf u, \bf v)=\check\nabla_{\bf u}\bf v - \nabla_{\bf u} \bf v.$$
I have also tried to rewrite the torsion tensor $Q$ in abstract index notation, which I also guess has the form:
$$ \Sigma^c_{a,b}u^a v^b = u^a \nabla_ a v^c - v^b\nabla_bu^c-[u,v]^c.$$
Thanks in advance for the help!
Note that \begin{align} [u,v]f &= u(vf)-v(uf) \\ &= \nabla_u(\nabla_v f) - \nabla_v(\nabla_u f) \\ &= u^a \nabla_a(v^b\nabla_b f) - v^b\nabla_b(u^a\nabla_a f) \\ &= u^a \nabla_a v^b\nabla_b f + u^a v^b\nabla_a\nabla_b f - v^b\nabla_bu^a\nabla_a f - v^bu^a\nabla_b\nabla_a f \\ &= (\nabla_u v)f - (\nabla_v u)f + u^a v^b(\nabla_a\nabla_b f - \nabla_b\nabla_a f) \end{align} So, if you define torsion as $$\Sigma(u,v) = \nabla_u v - \nabla_v u - [u,v]$$ then you get \begin{align} \Sigma_{ab}^c u^a v^b \nabla_c f = \Sigma(u,v)f &= (\nabla_u v)f - (\nabla_v u)f - [u,v]f \\ &= - u^a v^b(\nabla_a\nabla_b f - \nabla_b\nabla_a f) \\ &= u^a v^b(\nabla_b\nabla_a f - \nabla_a\nabla_b f). \end{align} That is $$ \Sigma_{ab}^c \nabla_cf = \nabla_b\nabla_a f - \nabla_a\nabla_b f, $$ which differs by a sign from your starting point. That's why your derivation went wrong.