Let $\Sigma$ be a compact $2$-manifold with riemannian metric $g$ and $f:\Sigma \to \mathbf{R}^n$ given locally by $f_1(x_1,x_2),\dots,f_n(x_1,x_2)$. Define $$ S(f,g) = -\frac{1}{2\pi\alpha'}\int_{\Sigma}\left(\sum_{i=1}^n\sum_{k,j=1}^2g^{jk}(x)\frac{\partial f_i}{\partial x_j}\frac{\partial f_i}{\partial x_k}\right)\Phi, $$ where $\Phi$ is the volume form.
Suppose that $g$ is the euclidean metric. If $f:\Sigma \to \mathbf{C}^n$ is given locally by $\phi_1(z),\dots,\phi_n(z)$ (using a single complex coordinate for $\Sigma$), the source I'm following says the above changes to $$ S(f,g) = -\frac{i}{2\pi\alpha'}\int \sum_{j=1}^n\left(\frac{\partial \phi_j}{\partial z}\frac{\partial \overline{\phi_j}}{\partial \bar z} + \frac{\partial \overline{\phi_j}}{\partial z}\frac{\partial \phi_j}{\partial \bar z}\right) dz \wedge d\bar z. $$
I get that in complex coordinates $\Phi = \frac{i}{2}dz \wedge d\bar z$ and $g^{jk} = 2$ if $j \neq k$ and $0$ otherwise, but I'm not sure how the $\overline{\phi_j}$ came up in the expression, and trying different guesses for what it should be didn't get me anywhere. What is going on here?
You should remember that the sum $ g^{jk} \; \partial_j f_i \; \partial_k f_i $ should be understood as $ g^{jk} \langle \partial_j f, \partial_k f \rangle $, where $ \langle , \rangle $ is the metric on the target space. In the case of $ \mathbb{R}^n $, it is just Euclidean. In the case the target space is $ \mathbb{C}^n $, the natural inner product is the hermitian inner product, $ \langle v, w \rangle = \sum_i \bar{v}_i w_i $.