Calculating monodromy

I'm sure there are better ways to do it, but as nobody has answered I hope this helps ...

I would proceed following the description you give in the following way, first the branch points are $0,-1$ and $\infty$, so if we shade the points that are sent to the upper plane under the map $$ z \to \frac{4z^2(z-1)^2}{(2z-1)^2} $$ we obtain a triangulation of the sphere were the ramification points will be placed in the vertex and the edges are sent to one of the three segments of the real line $(-\infty,-1),(-1,0),(0,\infty)$. Now give a number to each white triangle. A small loop around one of the branch points, say $\beta$ will come from disjoint loops say $\gamma_1, \gamma_2, \dots$ around some of the vertex, now if $\gamma_1$ cuts the triangles numbered $i_1,j_1,k_1, \dots$ (in counter clock wise order), $\gamma_2$ cuts the triangles numbered $i_2,j_2,k_2,\dots$ and so on then associate to the branch point $\beta$ the permutation $(i_1 j_1 k_1\dots)(i_2j_2k_2\dots)\dots$. In your example we have the following figure:

in this figure the two white dots are sent to $0$, the two filled ones to $-1$ and the cross (and $\infty$) to $\infty$ the preimage of a loop around $0$ will then be associated to the permutation $(14)(23)$ as small loops around the left white point cross the triangles 2 and 3 and a loop around the right white point crosses the triangles 1 and 4. Identically a loop around $-1$ with the permutation $(12)(34)$ and one around $\infty$ with $(13)(24)$. Renumbering the triangles in any way lead to conjugate triples.

If you have a branch of order two over a special point, the induced permutation is a two-cycle that switches the sheets connected to the ramified point, and if you have two order-two points over your branch point, you get a product of two disjoint 2-cycles. In $S_4$, there are only three elements of this form, which you listed in the statement of the question.

The remaining problem is to see how the permutations over two branch points relate to each other. The rule to remember is that the product of all three monodromies is trivial, since the composition of three loops yields a contractible loop. If two permutations are equal, then they multiply to identity. This is not allowed, because multiplying with the monodromy around the third point needs to be trivial, and that forces the third element to be trivial. The remaining option is that the three monodromies are matched with the three possible products of two transpositions in some way.

This answer is unique in the sense that the action of $S_4$ by renaming the numbers is transitive on the set of admissible assignments. Indeed, it factors through the permutation action of $S_3$ on the three group elements in question.