Let $f: S \to \Bbb R$ be a Morse function on a Riemann surface. Let $x\_0$ be a saddle point of $f$. Since $x\_0$ is a critical point of $f$, it makes sense to talk about the bilinear forms $f\_{z\overline{z}}(x\_0)$, $f\_{zz}(x\_0)$ and $f\_{\overline{z}\overline{z}}(x\_0)$ as components of the real Hessian of $f$ at $x\_0$ (if one only allows only complex coordinates).


Can $f$ be arranged so that the Hermitian form $f\_{z\overline{z}}$, which can be defined on all $S$, is arbitrarily small (compared to some fixed Hermitian metric) and $|f\_{zz}(x\_0)|$(as a number) is arbitrarily big in a fixed coordinate system around $x\_0$?



Special Morse function on a Riemann surface
First, let me point out that only the differential form $\newcommand{\pa}{\partial}$ $\newcommand{\bpa}{\bar{\partial}}$ $\pa\bpa f$ is well defined globally on $S$. Locally


$$ \pa\bpa f= f\_{z\bar{z}} dz\wedge d\bar{z}, $$


but the coordinate $z$ is only locally defined. If you choose different a different local coordinate $u$ you have an equality


$$ f\_{z\bar{z}} dz\wedge d\bar{z}= \pa\bpa f=f\_{u\bar{u}} du\wedge d\bar{u} \tag{A} $$


yet 


$$ f\_{z\bar{z}} \neq f\_{u\bar{u}}. $$


The equality (A) implies that 


$$ f\_{u\bar{u}}= f\_{z\bar{z}} \cdot \left| \frac{dz}{du} \right|^2. $$


On the other hand,


$$\frac{d}{du} =\frac{dz}{du}\frac{d}{dz} $$


so that


$$\frac{d^2}{du^2}= \frac{d^2 z}{du^2} \frac{d}{dz}+ \left(\frac{dz}{du}\right)^2\frac{d^2}{dz^2}. $$


Hence, if $z= au+bu^2+ O(u^3)$$, $$ a\in \mathbb{C}\setminus 0$$, we deduce


$$ f\_{u\bar{u}}= |a|^2 f\_{z\bar{z}}, $$


$$f\_{uu}= a^2 f\_{zz}. $$


In particular, at the critical point $x\_0$ we have


$$\left| \frac{f\_{uu}}{f\_{u\bar{u}}}\right|= \left| \frac{f\_{zz}}{f\_{z\bar{z}}}\right|. $$


This proves that you cannot find a holomorphic coordinate $u$ near $x\_0$ that makes $f\_{u\bar{u}}$ very small and $f\_{uu}$ very large.


**Update.** $\newcommand{\ii}{\boldsymbol{i}}$ If $z=x+\ii y$, then using the equalities


$$\pa\_z=\frac{1}{2}(\pa\_x-\ii\pa\_y),\;\;\pa\_{\bar{z}}=\frac{1}{2}(\pa\_x+\ii\pa\_y), $$


we deduce 


$\DeclareMathOperator{\tr}{tr}$ $\DeclareMathOperator{\Hess}{Hess}$


$$
f\_{z\bar{z}}=\frac{1}{4} (f\_{xx}+f\_{yy}) =\frac{1}{4}\tr \Hess(f), $$


$$ f\_{zz}=\frac{1}{4}( f\_{xx}+f\_{yy}+2\ii f\_{xy}). $$


As I mentioned in one of my comments, you can choose $f$ so that the partials $f\_{xx},f\_{yy}, f\_{xy}$ have any prescribed values at $x\_0$, but observe that


$$\left|\frac{f\_{zz}}{f\_{z\bar{z}}}\right|^2=\frac{ (\tr \Hess(f))^2+(f\_{xy})^2}{(\tr \Hess(f))^2}\geq 1. $$


