So, I am aware of the fact that if I do have a function like
$$f(z) = z^{1/2} + z^{1/3}$$
being $2, 3$ relatively primes, I have
$$\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$$
hence $6$ Riemann sheets.
What about having
$$g(z) = z^{1/2} + z^{1/4} + z^{1/3}$$
In this case, despite $2$ and $4$ share a divisor, $3$ and $4$ do not, hence I expect to have $24$ sheets, right? Or are they too many, and I just need $12$?
Also if
$$h(z) = z^{1/2} + z^{1/4}$$
then being $2$ and $4$ not relatively primes, I only have one Riemann sheet, right?
This question is more a sort of general doubt, so I just need a clarifying answer, not a very detailed treaties!
In order to answer such questions, first of all, one has to make sense of multivalued analytic functions and their Riemann surfaces. One formalism for this comes from algebraic topology as described below (see also this mathoverflow discussion; a more old-fashioned definition can be found for instance in "Complex Analysis" by Lars Ahlfors).
Let $U$ be a nonempty open connected subset of the complex plane (in your example, $U$ is the complement to $\{0\}$ in ${\mathbb C}$). Let $D\subset U$ be a connected subset and $\phi: D\to {\mathbb C}$ be a (single-valued!) analytic function. In you examples, $D$ can be taken to be a subset of the form $\{0< arg(z) < \alpha\}$ for a suitable $\alpha$. Assume that $\phi$ admits a multivalued analytic continuation (along paths), denoted $\Phi$, to the entire $U$. Let $p: X\to U$ be the universal covering, $\tilde\phi: \tilde{D}\to {\mathbb C}$ a lift of $\phi$. (Here $\tilde{D}$ is a connected component of $p^{-1}(D)$.) Then, by the monodromy principle, $\tilde\phi$ admits analytic continuation to the entire $X$ as a single valued function $\tilde\Phi: X\to {\mathbb C}$. (In a suitable sense, this continuation is a lift of $\Phi$. One can make this totally rigorous.) Now, let $G$ denote the group of covering transformations of $p: X\to U$. Let $H< G$ denote the subgroup consisting of elements $h\in G$ such that $$ \tilde\Phi\circ h= \tilde\Phi. $$ (One says that the function $\tilde\Phi$ is $H$-invariant.) Then $\tilde\Phi$ projects to a (single valued) analytic function
$$ \Psi: Y=X/H\to {\mathbb C} $$ $$ \Psi\circ q= \tilde\Phi, $$ where $q: X\to Y$ is the covering given by the quotient map $X\to X/H$. Then $Y$ is called the Riemann surface of $\phi$ (or $\tilde\phi$). In the case when the complement to $U$ is a finite subset and the covering map $Y\to U$ has finite degree one typically extends $Y$ to a compact Riemann surface by "filling in the punctures". Let's not worry about this part.
In general, this definition/construction is hard to implement (and one uses Riemann cuts instead), but in you examples, it is quite easy. Namely, the universal covering $p: X\to U= {\mathbb C}^*$ is the exponential map $$ p: w\mapsto exp(w), w\in {\mathbb C}=X. $$ The group $G$ is infinite cyclic, generated by
$$ \gamma: w\mapsto w+ 2\pi i. $$ Take now $\phi(z)= z^{1/2} + z^{1/4} + z^{1/3}$. This function lifts to the function $$ \tilde\Phi(w)= \exp(\frac{w}{2}) + \exp(\frac{w}{4}) + \exp(\frac{w}{3}) $$ on $X={\mathbb C}$. This function $\tilde\Phi$ is $H$-invariant for the subgroup $H< G$ generated by the translation $$ \gamma^{12}: w\mapsto w+ 12\cdot 2\pi i $$ and is not invariant under $\gamma^k,$ $k=4, 6$ (the maximal divisors of $12$ which are less than $12$). I will leave you to check this. Hence, $H=\langle \gamma^{12}\rangle$ is the required subgroup (that is, required by the definition I gave above). Its index in $G$ equals $12$: $$ |G: H|=12, $$ hence the covering $$ Y=X/H\to U $$ has degree 12. (That's something you learn in an algebraic topology class.) In Riemann's terminology, the Riemann surface of $\phi$ has 12 sheets.
As an aside, in this example, $Y$ is biholomorphic to ${\mathbb C}^*$, hence, you compactify it as the Riemann sphere.