I have been reading about Puiseux series in the context of the Newton–Puiseux algorithm for resolution of singularities of algebraic curves in $\mathbb{C}^2$. Given a curve $f(x,y)=0$ with $f$ a convergent power series and such that the curve has a singularity at the origin, the algorithm produces convergent Puiseux series $\varphi_1, \dotsc, \varphi_m$ verifying $f(x,y) = (y-\varphi_1(x))\dotsm(y-\varphi_m(x))$.
All of the books I have consulted start by constructing the Puiseux series as a formal power series and then invoking the Implicit Function Theorem to show that there exist analytic solutions to the problem, and therefore the constructed series must be convergent (in a small enough neighbourhood of the origin). Then they continue by studying the properties of the singularity using $\varphi_1, \dotsc, \varphi_m$ as parametrizations of the curve.
Here is where I have a problem. Say for instance that one of the resulting Puiseux series is $\varphi_1(x) = x^{3/2} + x^{7/4}$. This is not a well-defined complex function, is it? For $x=1$ we could have $x^{3/2}$ equal to $1$ or $-1$, and $x^{7/4}$ equal to $\pm1$ or $\pm i$, so there are eight different determinations of this particular Puiseux series. In general, if the Puiseux series has infinitely many terms, there would be infinitely many determinations.
As far as I understand, only one of these determinations of $\varphi_1$ is the actual function you want to consider, which is the function that the Implicit Function Theorem gives you — but because of the nature of the proof, you can't really know which one it is! It is not clear to me how do these books account for this problem, but their treatment of the series is as if they were well-defined functions (see for instance Brieskorn–Knörrer).
Best Answer
One standard way to bring actual functions in the picture is the following formulation of the existence + convergence results of Puiseux roots: "Given an irreducible power series $f \in \mathbb{C}[[x, y]]$ which is a Weirstrass polynomial in $y$ of degree $d$ (i.e. $f = y^d + \sum_{i=1}^d f_i(x)y^{d-i}$ with $f_i(x) \in x\mathbb{C}[[x]]$), there is $\phi(t) \in \mathbb{C}[[t]]$ such that $$f(t^d, y) = \prod_{i = 1}^d (y - \phi(\zeta^it))$$ where $\zeta$ is a $d$-th primitive root of $1$. In addition, and here is where you have actual functions, if $f$ is convergent, then so is $\phi$."
Added later (prompted by OP's comment about choices for $\phi$): in this formulation the uniqueness of Puiseux roots is the statement that if $\psi(t)$ is any other power series such that $y - \psi(t)$ divides $f(t^d, y)$, then $\psi(t) = \phi(\zeta^i t)$ for some $i$. In other words, there are precisely $d$, not infinitely many, Puiseux roots of $f$, namely $\phi(\zeta^it)$, $i = 1, \ldots, d$.
This is covered, e.g. in chapter one of Singularities of Plane Curves by Casas-Alvero.