Let $(M, h)$ be a Hermitian manifold. We then have a Riemannian metric $g=\Re(h)$ and a real-valued differential $2$-form $\omega = - \Im(h)$.
I want to see why locally we can write $$\omega = \frac{i}{2} \sum_{j,k} h_{jk} dz_j \wedge d\bar{z}_k$$ but I'm not sure how this expression is derived. If I have understood correctly $\omega$ is of type $(1,1)$? Starting from $\omega = - \Im(h)$ if we consider vectors $u, v$ where $u$ is $\mathbb{C}$-linear and $v$ $\mathbb{C}$-antilinear then can we expand $$\omega(u,v) = -\Im(h(u,v))$$ to get to this local expression?
A complex $k$ form $\omega$ on $M$ is said to be a $(p,q)$-form if $p + q = k$ and each $\omega_x : \Lambda^kT_xM \rightarrow \mathbb{C}$ (which is $\mathbb{R}$-multilinear) verifies for all $(u_1,\ldots,u_k) \in (T_xM)^k$ and all $z \in \mathbb{C}$, $$ \omega_x(zu_1,\ldots,zu_k) = z^q\overline{z}^p\omega_x(u_1,\ldots,u_k). $$ In particular, $\omega$ is a $(k,0)$ form, each $\omega_x$ is $\mathbb{C}$-antilinear and in your case, $$ \omega_x(zu,zv) = -\Im(h_x(zu,zv)) = -\Im(z\overline{z}h_x(u,v)) = -z\overline{z}\Im(h_x(u,v)) = z\overline{z}\omega_x(u,v). $$ It works because $z\overline{z} = |z|^2$ is real. Now, in a local holomorphic coordinates system $z_1,\ldots,z_n$, set $h_{j\overline{k}} = h\left(\frac{\partial}{\partial z_j},\frac{\partial}{\partial z_k}\right)$. It is a smooth function from an open subset of $M$ to $\mathbb{C}$ and for all vector fields $X = \sum_{j = 1}^n X_j\frac{\partial}{\partial z_j}$ and $Y = \sum_{j = 1}^n Y_j\frac{\partial}{\partial z_j}$, we have, \begin{align*} h(X,Y) & = h\left(\sum_{j = 1}^n X_j\frac{\partial}{\partial z_j},\sum_{k = 1}^n Y_k\frac{\partial}{\partial z_k}\right)\\ & = \sum_{j = 1}^n\sum_{k = 1}^n X_j\overline{Y_k}h\left(\frac{\partial}{\partial z_j},\frac{\partial}{\partial z_k}\right)\\ & = \sum_{j = 1}^n\sum_{k = 1}^n h_{j,\overline{k}}X_j\overline{Y_k}\\ & = \sum_{j = 1}^n\sum_{k = 1}^n h_{j,\overline{k}}(dz_j \otimes d\overline{z_k})(X,Y). \end{align*} Hence, $h = \sum_{j,k = 1}^n h_{j\overline{k}}dz_j \otimes d\overline{z}_k$.
Now, the conjugate symmetry of $h$ implies that for all $j,k$, $h_{k\overline{j}} = \overline{h_{j\overline{k}}}$. In other words, the matrices $(h_{j\overline{k}}(x))$ are Hermitian (and definite positive because $h$ is) so for all $j,k$, \begin{align*} \Im(h_{j\overline{k}}dz_j \otimes d\overline{z_k}) & = -\frac{i}{2}\left(h_{j\overline{k}}dz_j \otimes d\overline{z_k} - \overline{h_{j\overline{k}}dz_j \otimes d\overline{z_k}}\right)\\ & = -\frac{i}{2}\left(h_{j\overline{k}}dz_j \otimes d\overline{z_k} - h_{k\overline{j}}d\overline{z_j} \otimes dz_k\right). \end{align*} We deduce that, \begin{align*} \omega & = \sum_{j,k = 1}^n \frac{i}{2}\left(h_{j\overline{k}}dz_j \otimes d\overline{z_k} - h_{k\overline{j}}d\overline{z_j} \otimes dz_k\right)\\ & = \frac{i}{2}\sum_{j,k = 1}^n h_{j\overline{k}}dz_j \otimes d\overline{z_k} - \frac{i}{2}\sum_{j,k = 1}^n h_{k\overline{j}}d\overline{z_j} \otimes dz_k\\ & = \frac{i}{2}\sum_{j,k = 1}^n h_{j\overline{k}}dz_j \otimes d\overline{z_k} - \frac{i}{2}\sum_{j,k = 1}^n h_{j\overline{k}}d\overline{z_k} \otimes dz_j\\ & = \frac{i}{2}\sum_{j,k = 1}^n h_{j\overline{k}}(dz_j \otimes d\overline{z_k} - d\overline{z_k} \otimes dz_j)\\ & = \frac{i}{2}\sum_{j,k = 1}^n h_{j\overline{k}}dz_j \wedge d\overline{z_k}. \end{align*}