Is there a theory of NewtonPuiseux type expansions which works to parameterize singular surface germs $F(x,y,z) =0$? Ideally, each branch would be the image of map of the form the $x = u^m, y = v^n, z = \Sigma_{i + j > n} a_{i j} u^i v^j$ (after a linear change of the variables $x, y, z$).
I don't believe there is anything as general as that, but when your polynomial is over the complex numbers one can do the following. First shift and rotate coordinates so that you're working on a neighborhood of the origin where $F(0,0,0) = 0$ and $\partial_z^n F(0,0,0) = 0$ for some $n$. Then you can use the Weierstrass preparation theorem and (ignoring a nonvanishing factor) write $F(x,y,z)$ as $z^n + a_{n1}(x,y)z^{n1} + ... + a_0(x,y)$. If the discriminant of this polynomial, viewed as a function of $x$ and $y$ has normal crossings (i.e. is a monomial times a nonvanishing factor), one can use the JungAbhyankar theorem to get a factorization of the form $(z  b_1(x,y))...(z  b_n(x,y))$ where the $b_i$ are analytic in fractional powers of $x$ and $y$ as you want. Even if the discriminant is not normal crossings, you can first resolve the singularities of the discriminant to make it normal crossings and there are a few ways to do this. One way involves subdividing the $xy$ space into wedges, and on each wedge there is a way of first doing a coordinate change of the form $(x,y) > (x,y  g(x))$ then a "blowing up" coordinate change $(x,y) = (x',(x')^m y')$ and get the discriminant as you want. ($m$ may not be an integer, and also you might have to reverse the roles of the $x$ and $y$ variables). Then one proceeds as above. Of course the final result isn't as nice as you wanted, but this is in the nature of things. One can iterate the above to deal with analogues in higher dimensions, but the formulas get a lot more elaborate as one might guess. There's a paper by Parusinski "On the preparation theorem for subanalytic functions" that discusses a lot of these issues. 


If $k$ is an algebraically closed field of characteristic zero then the algebraic closure of $k((x))$ is $F = \cup k((x^{1/n}))$, the field of Puiseux series. In particular, the algebraic closure of $k(x)$ is contained in $F$. Now you want to look at the algebraic closure of $k(x,y)=k(x)(y)$, which is contained in the algebraic closure of $F(y)$, which in turn is contained in $\cup F((y^{1/n}))$. I think that gives you what you want. Edit: An element of $\cup F((y^{1/n}))$ is a power series in a (fixed) root of $y$ all of whose coefficients are powers series in roots of $x$ but it's not clear that one can bound $m$ such that all coefficients are in $k((x^{1/m}))$ for a fixed $m$. 

