Let consider $C \subset \mathbb{CP}^2$ defined by equation $F(x, y, z) = 0$.
By definition, the class group $\text{Cl}(C)$ is defined as a quotient by following exact sequence:
$$0 \to \mathbb{C}^\times \to K_C^\times \to \text{Div}(C) \to \text{Cl}(C) \to 0$$
Could anybody explain how explicitely the map $K_C^\times \to \text{Div}(C)$ from the multiplicative group of the function field to the divisors is given?
Since your variety happens to be a Riemann surface, note that $K_C$ consists of all meromorphic functions on $C$. Given a meromorphic function $f$ on $C$, we naturally get a divisor by writing down the zeroes and poles of $f$ with the corresponding orders: $$(f) = \sum_{p\in C} \text{ord}_p(f) [p],$$ where $\text{ord}_p(f)$ is the degree of the first coefficient of the Laurent series of $f$ at $p$.
Any divisor obtained this way is called a principal divisor, so the divisor group in some sense measures how far divisors can get from being principal.