Let $F$ be a (totally real if necessary) number field and $E$ a (totally imaginary if necessary) quadratic extension of $F$.
Why are half the places of $F$ split in $E$?
I mainly do not grasp the relation between splitting and the reciprocity laws, I would like to relate it to the Hilbert symbol $(a,b)_v$, so that the product formula $\prod_v (a,b)_v$ for suitable $a,b$ would give the result.
So, is there any mean to characterize the splitting places by a property like $(a,b)_v = 1$?
The answer to your first question is supplied by Kronecker's density theorem: in a finite extension $L/K$ of number fields with normal closure $N/K$, the Dirichlet density of primes that split completely in $L/K$ (or, equivalently, in $N/K$) is given by $1/(N:K)$. The proof is easy, but involves the fact that the Dedekind zeta function $\zeta_F(s)$ of number fields $F$ has the property that $(s-1)\zeta_F(s)$ has a finite nonzero limit as $s \to 1$. It is a special case of the better known Frobenius density theorem.
In the quadratic case $L = K(\sqrt{\alpha}\,)$, a prime ideal ${\mathfrak p}$ with odd norm splits if and only if $(\alpha/{\mathfrak p}) = 1$. You can express this Legendre symbol in terms of Artin symbols and then in terms of norm residue symbols, and the product formula allows you to write it as a product of symbols away from ${\mathfrak p}$; I do not know, however, what you actually mean by the second part of your question.