There are a few ways to generalize quadratic reciprocity. Let me describe for you how class field theory over $\mathbb{Q}$ works.
Let $K$ be a finite abelian extension of $\mathbb{Q}$ of degree $n$, Galois group $\Gamma$. One goal of class field theory is the following: give a general rule describing how prime numbers $p$ decompose into a product of primes in $\mathcal O_K$. It's cumbersome to give a succinct rule which describes how all primes split, so the usual workaround is to give a rule which applies to all primes except those in a finite $S$, usually containing all the ramified primes.
This generalizes quadratic reciprocity in the following way: let $K = \mathbb{Q}(\beta)$ for some $\beta \in \mathcal O_K$, and let $f \in \mathbb{Z}[X]$ be the minimal polynomial of $f$ over $\mathbb{Q}$. For almost all primes $p$, you have $(\mathcal O_K)_{(p)} = \mathbb{Z}_{(p)}[\beta]$, and so you can apply the following criterion: if you factor $f$ into a product of irreducibles $p_1^{e} \cdots p_g^{e}$ over $\mathbb{Z}/p\mathbb{Z}[X]$, then $e$ will be the ramification index of $p$ in $K$, the degree $f$ of any of the $p_i$ will be the inertia, and $p$ will split into $\frac{n}{ef}$ primes.
So splitting of primes is analogous to factoring certain polynomials over $\mathbb{Z}/p\mathbb{Z}[X]$. In particular, quadratic reciprocity deals with determining for which primes $p$ the polynomial $X^2 - q \in \mathbb{Z}/p\mathbb{Z}$ is irreducible for a fixed prime $q$ (possibly $q$ is a negative prime). In other words, quadratic reciprocity answers the question of which primes $p$ split in $\mathbb{Q}(\sqrt{q})$, or more generally which primes split in a quadratic extension of $\mathbb{Q}$.
Now, here is how class field theory gives you an algorithm for determining (with a finite number of exceptions) how primes split in $K$, or at least implies the existence of such an algorithm. Given a prime number $p$, fix a prime $\mathfrak p$ of $K$ lying over $p$. Remember that for $p$ unramified in $K$, the decomposition group $$\Gamma_{p} = \{ \sigma \in \Gamma : \sigma \mathfrak p = \mathfrak p \}$$ is cyclic, and it has a particularly nice generator, commonly denoted $(p, K/\mathbb{Q})$. It does not depend on the choice of $\mathfrak p$, because $K/\mathbb{Q}$ is abelian. It is called the Frobenius at $p$. The order $f = f_p$ of $(p,K/\mathbb{Q})$ is the inertial degree of $p$ in $K$, and $p$ splits into $\frac{n}{f}$ primes in $K$.
So given a divisor $g$ of $n$, we want an algorithm for determining which primes $p$ split into $g$ primes in $K$, i.e. which primes satisfy $g = \frac{n}{f_p}$.
The nonconstructive part of the proof is the Kronecker-Weber theorem. It says that there exists some integer $m$ such that $K$ is contained in $\mathbb{Q}(\zeta_m)$. Suppose we have found such an $m$. Let $L = \mathbb{Q}(\zeta_m)$, $G = \textrm{Gal}(L/\mathbb{Q})$, and $H = \textrm{Gal}(L/K)$. There is a canonical identification of $G$ with the group $(\mathbb{Z}/m\mathbb{Z})^{\ast}$, and under this identification we can think of $H$ as a subgroup of $(\mathbb{Z}/m\mathbb{Z})^{\ast}$. For $p$ not dividing $m$, the Frobenius $(p,L/\mathbb{Q})$ has the effect $\zeta_m \mapsto \zeta_m^p$. In particular, $(p,L/\mathbb{Q})$ can be identified with $p$ modulo $m$, and the inertial degree of $p$ in $L$ is then the multiplicative order of $p$ modulo $m$.
In general, if $\phi: A \rightarrow B$ is a homomorphism of finite groups, and $a \in A$, then the order of $\phi(a)$ is the smallest number $f$ such that $a^f \in \textrm{Ker } \phi$. Now, you can check that the restriction of $(p,L/\mathbb{Q})$ to an automorphism of $K$ is exactly $(p,K/\mathbb{Q})$. Therefore, the inertial degree of $(p,K/\mathbb{Q})$ is the smallest number $f$ such that $p^f$ is congruent modulo $m$ to a member of $H$.
So to summarize, here is how to determine how primes split in $K$:
Find an $m$ such that $K \subseteq \mathbb{Q}(\zeta_m)$.
Identify the Galois group of $\mathbb{Q}(\zeta_m)$ with $(\mathbb{Z}/m\mathbb{Z})^{\ast}$, and identify $H = \textrm{Gal}(\mathbb{Q}(\zeta_m)/\mathbb{Q})$ as a subgroup of $(\mathbb{Z}/m\mathbb{Z})^{\ast}$.
With the exception of the prime factors of $m$, the primes $p$ which split into $g$ factors in $K$ are exactly those primes for which $p+H$ has order $\frac{n}{g}$ in $G/H$.
