It's nearly the same story as when you construct Riemann surface for a function $\sqrt{z^2}$: it's two $\mathbb C$s glued along $0$, and going around the ramification point will never bring you from one sheet to another — you have to pass through it to change sheets. The difference between, say, yours $z^{1/2}+z^{1/4}$ and $z^{1/2}+z^{1/5}$ is that 2 and 5 are relatively prime, while 2 and 4 are not. What's essential in understanding monodromy of multivalued functions is that when you focus on a particular ramification point, gluing is controlled by homomorphism $\mathbb Z \to Sym(sheets)$, where $\mathbb Z$ is fundamental group of little disc minus point. But the image is restricted by the form of a function — sheets cannot be shuffled arbitrarily by going around, and possible permutations are restricted by the fact that you cannot, say, pass from $(z^{1/2}, z^{1/4})$ sheet to $(z^{1/2}, iz^{1/4})$ one by traversing along any path missing zero — the parity of square and quartic roots of unity should remain same! If root powers were 2 and 5, then corresponding shuffling group contains a copy of $\mathbb Z / 2 \mathbb Z \times \mathbb Z / 5 \mathbb Z$, but this one equals $\mathbb Z / 10 \mathbb Z$ because 2 and 5 are relatively prime, and loop going around origin corresponds to element $(1, 1)$ which is exactly its generator. And in your case to traverse along all 8 sheets in one sweep you'd need the group $\mathbb Z / 2 \mathbb Z \times \mathbb Z / 4 \mathbb Z$ to be generated by one element, and it's impossible.
There's nice book about cuts, branches (and using them to prove Abel's theorem) in PIY (prove it yourself) style: V. B. Alekseev, Abel’s Theorem in Problems & Solutions
