Though I don't have it in front of me for the details, check out chapter I of Hartshorne...it's much less terrifying than the rest of the book is for non-algebraic geometry people.  Given a function field of transcendent degree one, you can take all the DVR's with field of fractions that field, and they'll form the points of a Riemann surface (actually, smooth projective algebraic curve over an algebraically closed field).

As for polynomials, yes, any such field is the field of fractions of a field of the form $k[x,y]/(f)$ but not uniquely, as KConrad pointed out, for rather simple reasons.  But also in Hartshorne chapter I, it's prove that every variety is birational to a hypersurface in affine space, and function fields are the same as birational classes of varieties.  As for the singularities that occur, which will occur generically, because most curves of large genus don't embed into the plane smoothly, you can just compute the normalization of the curve.

As Qiaochu said, there's [this question][1] which talks about the function field and the genus, and as for generators, the fact that things are birational to singular plane curves will tell you that you can always choose two generators to make things work out.

Now, every variety has a function field, as for every function field having a variety, I believe it's true, but don't have a proof off the top of my head (though that might be that it's 1:30 am and it's obvious when I'm awake).  As for the proposed anti-equivalence, you'll need to make it dominant rational maps, and then I believe it's true.  Dominant means that the image is dense, and if it fails, then the image might be in the exceptional locus of the next morphism, and composition doesn't work out, because then you don't have a rational map, you have the empty map.

I think I covered most of your questions...but really, chapter I of Hartshorne would be a good read for you.

  [1]: http://mathoverflow.net/questions/152/how-do-you-see-the-genus-of-a-curve-just-looking-at-its-function-field