[proofplan]
We prove the identity by expanding the cyclic curvature sum using the stated curvature convention. The commutator part of the curvature is rewritten with the torsion identity $\nabla_v w - \nabla_w v = [v,w] + T(v,w)$. The first-order torsion terms become the cyclic sum of covariant derivatives of $T$, while the remaining connection and bracket terms reduce to the cyclic sum of $T(T(\cdot,\cdot),\cdot)$ by skew-symmetry of torsion and the Jacobi identity for the Lie bracket.
[/proofplan]
[step:Introduce cyclic notation and record the torsion identities]
For any trilinear expression $A$ in $u,v,w$, define its cyclic sum by
\begin{align*}
\mathfrak S A(u,v,w) := A(u,v,w) + A(v,w,u) + A(w,u,v).
\end{align*}
With this notation, the desired identity is
\begin{align*}
\mathfrak S R(u,v)w = \mathfrak S (\nabla_uT)(v,w) + \mathfrak S T(T(u,v),w).
\end{align*}
The torsion tensor is skew-symmetric. Indeed, using the skew-symmetry of the Lie bracket,
\begin{align*}
T(v,u) = \nabla_v u - \nabla_u v - [v,u] = -\nabla_u v + \nabla_v u + [u,v] = -T(u,v).
\end{align*}
Also, the defining formula for $T$ gives
\begin{align*}
\nabla_v w - \nabla_w v = [v,w] + T(v,w).
\end{align*}
[/step]
[step:Expand the cyclic curvature sum]
Using the curvature convention in the statement,
\begin{align*}
\mathfrak S R(u,v)w = \mathfrak S\left(\nabla_u\nabla_v w - \nabla_v\nabla_u w - \nabla_{[u,v]}w\right).
\end{align*}
In the second second-order term, cyclic relabelling gives
\begin{align*}
\mathfrak S \nabla_v\nabla_u w = \mathfrak S \nabla_u\nabla_w v.
\end{align*}
Therefore
\begin{align*}
\mathfrak S R(u,v)w = \mathfrak S\left(\nabla_u(\nabla_v w - \nabla_w v) - \nabla_{[u,v]}w\right).
\end{align*}
Substituting $\nabla_v w - \nabla_w v = [v,w] + T(v,w)$ yields
\begin{align*}
\mathfrak S R(u,v)w = \mathfrak S \nabla_u[v,w] - \mathfrak S\nabla_{[u,v]}w + \mathfrak S\nabla_u(T(v,w)).
\end{align*}
[/step]
[step:Separate the covariant derivatives of torsion]
By the definition of the covariant derivative of the tensor $T$,
\begin{align*}
\nabla_u(T(v,w)) = (\nabla_uT)(v,w) + T(\nabla_u v,w) + T(v,\nabla_u w).
\end{align*}
Applying this identity inside the cyclic sum gives
\begin{align*}
\mathfrak S R(u,v)w = \mathfrak S(\nabla_uT)(v,w) + \mathfrak S T(\nabla_u v,w) + \mathfrak S T(v,\nabla_u w) + \mathfrak S\nabla_u[v,w] - \mathfrak S\nabla_{[u,v]}w.
\end{align*}
It remains to identify the last four cyclic sums with $\mathfrak S T(T(u,v),w)$.
[/step]
[step:Collect the remaining terms into the quadratic torsion contribution]
Expand the quadratic torsion term using $T(u,v)=\nabla_u v-\nabla_v u-[u,v]$:
\begin{align*}
\mathfrak S T(T(u,v),w) = \mathfrak S T(\nabla_u v,w) - \mathfrak S T(\nabla_v u,w) - \mathfrak S T([u,v],w).
\end{align*}
Since $T$ is skew-symmetric, cyclic relabelling gives
\begin{align*}
-\mathfrak S T(\nabla_v u,w) = \mathfrak S T(v,\nabla_u w).
\end{align*}
Next, expanding $T([u,v],w)$ gives
\begin{align*}
-T([u,v],w) = -\nabla_{[u,v]}w + \nabla_w[u,v] + [[u,v],w].
\end{align*}
Taking cyclic sums,
\begin{align*}
-\mathfrak S T([u,v],w) = -\mathfrak S\nabla_{[u,v]}w + \mathfrak S\nabla_w[u,v] + \mathfrak S[[u,v],w].
\end{align*}
The cyclic derivative term satisfies $\mathfrak S\nabla_w[u,v] = \mathfrak S\nabla_u[v,w]$ by relabelling. The cyclic bracket term vanishes by the Jacobi identity for the Lie bracket:
\begin{align*}
\mathfrak S[[u,v],w] = 0.
\end{align*}
Therefore
\begin{align*}
\mathfrak S T(T(u,v),w) = \mathfrak S T(\nabla_u v,w) + \mathfrak S T(v,\nabla_u w) + \mathfrak S\nabla_u[v,w] - \mathfrak S\nabla_{[u,v]}w.
\end{align*}
[guided]
The goal of this step is to recognize the leftover terms from the curvature expansion as the nonlinear torsion expression. Start with the displayed cyclic term
\begin{align*}
\mathfrak S T(T(u,v),w).
\end{align*}
The inner torsion is
\begin{align*}
T(u,v)=\nabla_u v-\nabla_v u-[u,v],
\end{align*}
so bilinearity of $T$ gives
\begin{align*}
\mathfrak S T(T(u,v),w) = \mathfrak S T(\nabla_u v,w) - \mathfrak S T(\nabla_v u,w) - \mathfrak S T([u,v],w).
\end{align*}
We now rewrite the second term. Because $T$ is skew-symmetric, $T(a,b)=-T(b,a)$ for all complex vector fields $a,b$. Hence
\begin{align*}
-T(\nabla_v u,w)=T(w,\nabla_v u).
\end{align*}
After applying the cyclic sum and relabelling the cyclic variables, this becomes
\begin{align*}
-\mathfrak S T(\nabla_v u,w)=\mathfrak S T(v,\nabla_u w).
\end{align*}
It remains to rewrite the bracket term. Applying the definition of torsion to the pair $[u,v]$ and $w$ gives
\begin{align*}
T([u,v],w)=\nabla_{[u,v]}w-\nabla_w[u,v]-[[u,v],w].
\end{align*}
Multiplying by $-1$ and summing cyclically,
\begin{align*}
-\mathfrak S T([u,v],w) = -\mathfrak S\nabla_{[u,v]}w + \mathfrak S\nabla_w[u,v] + \mathfrak S[[u,v],w].
\end{align*}
The term $\mathfrak S\nabla_w[u,v]$ is the same cyclic sum as $\mathfrak S\nabla_u[v,w]$, because the triples $(w,u,v)$ and $(u,v,w)$ run through the same cyclic ordering. Finally,
\begin{align*}
\mathfrak S[[u,v],w]=[[u,v],w]+[[v,w],u]+[[w,u],v]=0
\end{align*}
by the Jacobi identity for the Lie bracket.
Substituting these two simplifications into the expansion of $\mathfrak S T(T(u,v),w)$ gives
\begin{align*}
\mathfrak S T(T(u,v),w) = \mathfrak S T(\nabla_u v,w) + \mathfrak S T(v,\nabla_u w) + \mathfrak S\nabla_u[v,w] - \mathfrak S\nabla_{[u,v]}w.
\end{align*}
This is exactly the leftover expression obtained after separating off the covariant derivatives of $T$.
[/guided]
[/step]
[step:Combine the identities and restrict to type components]
Substituting the identity from the previous step into the curvature expansion gives
\begin{align*}
\mathfrak S R(u,v)w = \mathfrak S(\nabla_uT)(v,w) + \mathfrak S T(T(u,v),w).
\end{align*}
Writing out the cyclic sums is precisely
\begin{align*}
\mathcal R(u,v,w) = (\nabla_uT)(v,w) + (\nabla_vT)(w,u) + (\nabla_wT)(u,v) + T(T(u,v),w) + T(T(v,w),u) + T(T(w,u),v).
\end{align*}
This proves the asserted identity for the complexified tangent bundle.
For the Chern connection on $T^{1,0}X$, the same computation applies to the complex-linear extension of the connection and bracket on the relevant complexified bundles. Since every term in the identity is tensorial and respects the chosen type projection or restriction, projecting the established complexified identity to the desired type component gives the stated Chern-connection form.
[/step]
