I have recently asked a question in a similar vein: What makes Geometric CFT easier than CFT?

but I'm afraid I wasn't quite ripe to ask it yet. I have since consulted with the following sources:

http://jmilne.org/math/Books/ADTnot.pdf, http://math.stanford.edu/~conrad/249BPage/handouts/geomcft.pdf and http://arxiv.org/abs/hep-th/0512172

As well as some sources refreshing my memory on classic CFT:

http://people.maths.ox.ac.uk/gounelas/projects/bmo.pdf and http://www.math.dartmouth.edu/~trs/expository-papers/tex/CFT.pdf among others.

My motivation for studying geometric class field theory was first and foremost to solidify my understanding of classic class field theory. I thought that perhaps the geometric intuition will shed some light on the rather massive apparatus of CFT.

In order to be concrete, I will specify a version of geometric CFT:

Theorem: Let $C$ be a smooth projective, geometrically irreducible curve over a finite field $k$, and let $K$ be its function field. Then the (Artin reciprocity) map $\Phi_K:Div(C)\rightarrow \pi_1^{ab}(C)$ given by $p\mapsto Frob_p$ factors through $Pic(C)$, and induces an isomorphism between the profinite completion of $Pic(C)$ and $\pi_1^{ab}(C)$.

This appears in Toth's master thesis (linked above) as Theorem 1.1.4. In order to understand this as analogous to classic CFT, one need only notice that $K^{\times}\backslash \mathbb{I}_K/ \prod _{p \mbox{ closed point in } C}\hat{O_p} $ (where $\mathbb{I}_K$ is the ideles, and $\hat{O_p}$ is the completion of the stalk at the closed point $p$) is isomorphic to $Pic(C)$.

In other words the theorem above can be viewed as analogous to the adelic point of view of CFT (i.e. the isomorphism between the profinite completions of $K^{\times}\backslash \mathbb{I}_K/\prod_v O_v^{\times}$ with $Gal(K^{ab}/K)$, where $K$ is a number field).

###Question

How does one understand geometric CFT (as described in the theorem above) in terms of modularity results? In other words, what is the analogous statement to the fact that (up to finitely many primes) the splitting of primes in an abelian extension of $\mathbb{Q}$ is determined by what those primes are conjugate to modulo some conductor? Is there a geometric intuition behind the analogous statement in geometric CFT?