I'm not sure how to answer the question about intuition. In regards to your second question, the Tannakian formalism works over any field and in fact works for more general categories than Rep(G). A category C is called a neutral Tannakian category over k (k a field) if it is a rigid abelian tensor category together with a k-linear tensor exact functor from C to k-Vect. This latter functor is known as the fiber functor (and is just the forgetful functor referenced in the original post when C = Rep(G)). In this case, there's a theorem that says any neutral Tannakian category is equivalent to the category of representations of an affine group scheme G, where G is given by the tensor automorphisms of the fiber functor.
One of the original references for this is a 1982 paper by Deligne and Mumford Milne titled 'Tannkian Categories.' Deligne also has an article in the more recent FGA explained.
My understanding is that sometimes k can even be replaced by an arbitrary commutative ring. I don't know general conditions under which this holds. However, a great example of the Tannakian formalism in action where k is a general commutative ring is the Mirkovic-Vilonen paper on the geometric Satake correspondence, which they prove in great generality (Ginzburg also has a nice paper on this topic, but only in characteristic zero). The Mirkovic-Vilonen paper can be found here http://arxiv.org/abs/math/0401222
In this paper, they prove that their fiber functor is represented by a k-module which is free over k (k any commutative ring here) and hence the Tannakian formalism still works. I don't know if this condition is also a necessary condition.
There's also a version of the Tannakian formalism over a scheme, so to speak. If we view a group G as a principal G-bundle on a point and the category of vector spaces as the category of vector bundles on a point, then we can try to generalize to a scheme. Namely, if we fix a G-bundle on a scheme X, then this induces an exact tensor functor from Rep(G) to the category of vector bundles on X using the associated bundle construction. It turns out that the converse holds: given an exact tensor functor from Rep(G) to vector bundles on X, this is equivalent to giving a G-bundle on X. There's a proof of this fact in a set of notes on the webpage for the seminar that Dennis Gaitsgory is currently running.