Let $f$ be a map between compact Riemann surface.
Let $e_P(f)$ be the ramification degree of $f$ at point $P$.
Let $\nu_P(f)$ be the order of $f$ at point $P$, meaning that the lowest term of the Laurent series of $f$ has order $\nu_P(f)$.
I think there is some connection between this two. For example, I feel that when $\nu_P(f)>0$, $e_P(f)=\nu_P(f)$ and when $\nu_P(f)<0$, $e_P(f)=-\nu_P(f)$, am I right?
I assume by ramification degree you mean multiplicity (what power of $n$ it locally looks like). The fact is that if $X$ is a Riemann surface and $f:X\to\mathbb{C}$ is meromorphic then $\text{mult}_p(F)=\text{ord}_p(f-f(p))$ if $p$ is not a pole of $f$ and $\text{mult}_p(F)=-\text{ord}_p(f)$ if $p$ is a pole of $f$ where $F$ is the induced map $\mathbb{C}\to\mathbb{P}^1$.
This is fairly easy to check. For example, by definition of multiplicity we can find a centered chart $(U,\varphi)$ at $p$ such that $\psi\circ F\circ\varphi^{-1}=z^{\text{mult}_p(F)}$ where $\psi$ is the standard centered chart at $\infty$ on $\mathbb{P}^1$. Now, the two functions $F\circ\varphi^{-1}$ and $f\circ\varphi^{-1}$ agree on a punctured neighborhood of $p$ and thus must have the same order at that point. Thus, $\text{ord}_p(f)=\text{ord}_p(F)$. But, $\psi$ has order $-1$ at infinity and so $\text{mult}_p(F)=\text{ord}_p(\psi\circ F\circ\varphi^{-1})=-\text{ord}_p(F)=\text{ord}_p(f)$.