The question seems to have been edited. I saw the query about $\sqrt+(z-2)^$. I'll do the revised version afterward. The branch points are where either of the two radicals behaves strangely (has fewer than the expected number of distinct roots), namely, 0 and 2.
Over every point other than 0, there are two branches of the square root. To make two single-valued square roots, "cutting" the plane along any ray from 0 to infinity works.
Over everypoint other than 2, there are three branches of that cube root. To make three single-valued cube roots, cut the plane along any ray from 2 to infinity.
To take care of both radicals simultaneously, basically we need two rays, one from 0 to infinity, the other from 2 to infinity. We could take a ray that starts at 0, passes through 2, and goes to infinity, but this is a slightly weird choice, and doesn't really help anything much.
The Riemann surface has to be/is a "ramified covering" of the plane on which the function is well-defined. This requires a six-sheeted covering to incorporate the $2\cdot 3$ sheets needed by the square root and the cube root.
To determine the genus via the Riemann-Hurwitz formula, observe that the point at infinity has six-fold ramification, giving one point with ramification degree 6. There are 3 different points with ramification degree 2, and 2 points with ramification degree 3. Thus, by Riemann-Hurwitz, $$ 2 - 2g' = 6(2 - 2g) - \sum_P (e_P-1) = 12 - ((6-1)+3(2-1)+2(3-1)) = 12 - 12$$ Thus, the genus of the cover is $g'=1$. [Edited: garbled Riemann-Hurwitz!]
With $\sqrt+(z-4)^$, there are 8 sheets altogether. The point at infinity behaves a little differently, since the cover on which $(z-4)^$ exists near infinity already has a $\sqrt$ there. Thus, there are two points lying over the point at infinity, each with ramification degree 4. There are 4 points lying over 0, each with ramification 2, and 2 points over 4, each of ramification 4. Riemann-Hurwitz is $$ 2-2g' = 8(2 - 2g) - \sum_P(e_P-1) = 16 - (2(4-1)+4(2-1)+2(4-1)) = 16 - 16$$ so the genus of the cover is $g'=1$.