A Hermitian metric $h$ on a complex manifold $X$ is Kähler if the associated $2$-form $\omega=\mathrm{Im} (h)$ is closed.
This condition is trivial on compact Riemann surface, implying that every Hermitian metric on a compact Riemann surface is Kähler.
An equivalent condition for a metric to be Kähler is that this metric "osculates to order 2" the standard metric, meaning that the components of the metric satisfy
$$ h_{ij}(z) = \delta_{ij} + O(|z|^2). $$
The proof of the fact that $d\omega=0$ implies this condition that most books do (for example, Griffiths-Harris or Huybrechts) is clear when $\dim X\geq 2$. However, the proof given in those books does not work for a Riemann surface since the condition $d\omega=0$ is vacuous in that case.
My question is how does one prove that, if $h$ is a Hermitian metric on a Riemann surface, that locally can be written as
$$ h=h(z) dz\otimes d\bar{z}, $$
then $h(z)=1+O(|z|^2)$.
You just follow the usual proof, but it's easier. $h$ is a positive (real analytic?) function and you write $h\,dz\otimes d\bar z = \phi\otimes\bar\phi$. Without loss of generality, we assume $h(0)=1$. Then the $(1,0)$-form $\phi$ is given by $$\phi(z) = \left(1+az+O(|z|^2)\right)dz.$$ Set $w=z+\frac a2 z^2 + O(|z|^3)$, and note that $$dw = \left(1+az+O(|z|^2)\right)dz,$$ so the $(1,0)$-form $\phi = \left(1+O(|w|^2)\right) dw$, as needed. The Kähler condition is totally redundant.