[proofplan] We work in the holomorphic coordinate frame $e_i=\partial/\partial z_i$ and use the defining properties of the Chern connection: its $(0,1)$-part is the holomorphic structure and it is compatible with the Hermitian metric. The first property forces the connection matrix in this frame to have type $(1,0)$, while metric compatibility determines it uniquely as $h^{-1}\partial h$. The curvature is then computed from the local matrix formula for a connection, and its $(1,1)$-component is obtained by applying $\bar\partial$ to the connection matrix. Finally, expanding the resulting matrix-valued form and converting from $d\bar z_l\wedge dz_k$ to $dz_k\wedge d\bar z_l$ gives the displayed component sign. [/proofplan]

[step:Use the holomorphic frame to isolate the connection matrix of type $(1,0)$] On the chart $U$, define the local holomorphic frame \begin{align*} e_i:U\to T^{1,0}X|_U \end{align*} by $e_i=\partial/\partial z_i$. By the stated convention for connection matrices, there are complex-valued $1$-forms $\theta_i^p$ on $U$ such that \begin{align*} \nabla^C e_i=\sum_p \theta_i^p\otimes e_p. \end{align*} Let $\theta=(\theta_i^p)$ denote this matrix of $1$-forms. Because $\nabla^C$ is the Chern connection, its $(0,1)$-part equals the holomorphic structure $\bar\partial$ on $T^{1,0}X$. Since each $e_i$ is a holomorphic local section, $\bar\partial e_i=0$. Therefore the $(0,1)$-part of $\nabla^C e_i$ vanishes for every $i$, and each $\theta_i^p$ has type $(1,0)$. Hence $\theta$ is a matrix of $(1,0)$-forms. [/step]

[step:Derive the connection matrix from metric compatibility]For indices $i$ and $q$, define the metric coefficient \begin{align*} h_{i\bar q}:U\to \mathbb{C} \end{align*} by $h_{i\bar q}=h(e_i,e_q)$. Let $V$ be a smooth vector field of type $(1,0)$ on $U$. Metric compatibility of $\nabla^C$ gives \begin{align*} V(h_{i\bar q})=h(\nabla^C_V e_i,e_q)+h(e_i,\nabla^C_{\bar V}e_q). \end{align*} The second term vanishes because the $(0,1)$-part of the Chern connection is $\bar\partial$ and $e_q$ is holomorphic, so $\nabla^C_{\bar V}e_q=0$. Using $\nabla^C e_i=\sum_p \theta_i^p\otimes e_p$, we obtain \begin{align*} V(h_{i\bar q})=\sum_p \theta_i^p(V)h_{p\bar q}. \end{align*} Since this identity holds for every $(1,0)$ vector field $V$, it is the matrix identity \begin{align*} \partial h_{i\bar q}=\sum_p \theta_i^p h_{p\bar q}. \end{align*} Multiplying by the inverse matrix $h^{-1}=(h^{p\bar q})$ gives \begin{align*} \theta_i^p=\sum_q h^{p\bar q}\partial h_{i\bar q}. \end{align*} Thus, in matrix notation with the convention of the statement, \begin{align*} \theta=h^{-1}\partial h. \end{align*}[/step]

[guided]The Chern connection is characterized by two conditions, and in this holomorphic frame each condition has a direct algebraic consequence. First, because $e_i=\partial/\partial z_i$ is a holomorphic frame vector, the condition $(\nabla^C)^{0,1}=\bar\partial$ gives \begin{align*} (\nabla^C)^{0,1}e_i=\bar\partial e_i=0. \end{align*} Therefore the connection forms $\theta_i^p$ have no $(0,1)$-part; they are $(1,0)$-forms. Now we use metric compatibility to identify those $(1,0)$-forms. Define \begin{align*} h_{i\bar q}:U\to \mathbb{C} \end{align*} by $h_{i\bar q}=h(e_i,e_q)$. If $V$ is a smooth vector field of type $(1,0)$, compatibility of $\nabla^C$ with $h$ says \begin{align*} V(h(e_i,e_q))=h(\nabla^C_V e_i,e_q)+h(e_i,\nabla^C_{\bar V}e_q). \end{align*} The term involving $\nabla^C_{\bar V}e_q$ is zero because $e_q$ is holomorphic and the $(0,1)$-part of $\nabla^C$ is exactly $\bar\partial$. Thus \begin{align*} V(h_{i\bar q})=h(\nabla^C_V e_i,e_q). \end{align*} Using the connection-matrix convention \begin{align*} \nabla^C e_i=\sum_p \theta_i^p\otimes e_p, \end{align*} we get \begin{align*} \nabla^C_V e_i=\sum_p \theta_i^p(V)e_p. \end{align*} Substituting this into the metric-compatibility identity gives \begin{align*} V(h_{i\bar q})=\sum_p \theta_i^p(V)h(e_p,e_q)=\sum_p \theta_i^p(V)h_{p\bar q}. \end{align*} Because this holds for every $(1,0)$ vector field $V$, the corresponding identity of $(1,0)$-forms is \begin{align*} \partial h_{i\bar q}=\sum_p \theta_i^p h_{p\bar q}. \end{align*} Finally, the Hermitian matrix $h=(h_{i\bar q})$ is positive definite at each point, so it is invertible. Let $h^{-1}=(h^{p\bar q})$ be its inverse, characterized by \begin{align*} \sum_q h^{p\bar q}h_{r\bar q}=\delta_r^p. \end{align*} Multiplying the preceding identity by $h^{p\bar q}$ and summing over $q$ isolates the coefficient $\theta_i^p$: \begin{align*} \theta_i^p=\sum_q h^{p\bar q}\partial h_{i\bar q}. \end{align*} This is exactly the local matrix formula \begin{align*} \theta=h^{-1}\partial h. \end{align*}[/guided]

[step:Compute the curvature matrix and extract its $(1,1)$-part] The curvature of a connection with local matrix $\theta$ is represented by the matrix-valued $2$-form \begin{align*} \Theta^C=d\theta+\theta\wedge\theta. \end{align*} Since every entry of $\theta$ has type $(1,0)$, the form $\partial\theta$ has type $(2,0)$, the form $\bar\partial\theta$ has type $(1,1)$, and $\theta\wedge\theta$ has type $(2,0)$. Therefore the $(1,1)$-component of the Chern curvature matrix is \begin{align*} (\Theta^C)^{1,1}=\bar\partial\theta. \end{align*} The Chern curvature is of type $(1,1)$, so the curvature matrix itself is \begin{align*} \Theta^C=\bar\partial\theta=\bar\partial(h^{-1}\partial h). \end{align*} [/step]

[step:Expand the matrix formula to obtain the coefficient sign] From the formula for $\theta_i^p$, define the coefficient functions \begin{align*} A_{ik}^p:U\to \mathbb{C} \end{align*} by \begin{align*} A_{ik}^p=\sum_q h^{p\bar q}\partial_k h_{i\bar q}. \end{align*} Then \begin{align*} \theta_i^p=\sum_k A_{ik}^p\,dz_k. \end{align*} Applying $\bar\partial$ gives \begin{align*} (\Theta^C)_i^p=\bar\partial\theta_i^p=\sum_{k,l}\partial_{\bar l}A_{ik}^p\,d\bar z_l\wedge dz_k. \end{align*} Since $1$-forms anticommute under the wedge product, \begin{align*} d\bar z_l\wedge dz_k=-dz_k\wedge d\bar z_l. \end{align*} Therefore \begin{align*} (\Theta^C)_i^p=-\sum_{k,l}\partial_{\bar l}A_{ik}^p\,dz_k\wedge d\bar z_l. \end{align*} Using the definition \begin{align*} R^C(e_i)=\sum_{p,k,l}R^p{}_{i k\bar l}\,dz_k\wedge d\bar z_l\otimes e_p, \end{align*} we read off \begin{align*} R^p{}_{i k\bar l}=-\partial_{\bar l}A_{ik}^p=-\partial_{\bar l}\left(\sum_q h^{p\bar q}\partial_k h_{i\bar q}\right). \end{align*} If instead the coefficients are read in the order $d\bar z_l\wedge dz_k$, the preceding computation gives the same coefficient before commuting the wedge factors, so the sign is reversed. This proves all asserted formulas. [/step]