Completely explicitly, let $G$ be the Galois group of the extension $K/\mathbb{Q}$. So $G$ is a quotient of the absolute Galois group $G_{\mathbb{Q}}$ of $\mathbb{Q}$. Then the product runs over the non-trivial irreducible complex representations of the absolute Galois group of $\mathbb{Q}$ that factor through $G$, i.e. over the non-trivial characters of $G$, thought of as characters of $G_{\mathbb{Q}}$. The zeta function of $\mathbb{Q}$ corresponds to the trivial character. Thus, in the example $K=\mathbb{Q}(\sqrt{3})$ the product only consists of one term, corresponding to the non-trivial character of $G\cong\mathbb{Z}/2\mathbb{Z}$, the so-called quadratic character attached to $K$.
This works in much greater generality: for any base field instead of $\mathbb{Q}$ and for arbitrary Galois groups, not necessarily abelian. The Dirichlet $L$-functions are then replaced by Artin $L$-functions.
Edit:
Here is an explanation of why the character is what I claim it is: first, embed $K$ into a cyclotomic field. In the particular example $K=\mathbb{Q}(\sqrt{3})$, we have $K\subset F=\mathbb{Q}(\mu_{12})$. The Galois group of $F/\mathbb{Q}$ is isomorphic to $(\mathbb{Z}/12\mathbb{Z})^\times\cong \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$. Once we fix a primitive 12-th root of unity $\zeta$, the isomorphism is given by $\sigma\mapsto a$ where $\sigma(\zeta) = \zeta^a$. You can explicitly write down the 4 Dirichlet characters of $(\mathbb{Z}/12\mathbb{Z})^\times$ and I will leave this to you. Now, you should identify the subgroup of $(\mathbb{Z}/12\mathbb{Z})^\times$ that corresponds to the field $K$ (notice that $F$ has three quadratic subfields: $\mathbb{Q}(\sqrt{\pm3})$ and $\mathbb{Q}(\sqrt{-1})$) and you will find that out of the three non-trivial Dirichlet characters you just write down, exactly one will factor through this subgroup. That's the character you are looking for. No serious class field theory comes into this, really, nothing beyond Kronecker-Weber.