If I understand your question correctly, the "complex sphere" is the same as the Riemann sphere, or the one point compactification of $\mathbb C$.

There is a well-developed theory of questions like this. Some of the keywords are Jenkins-Strebel differential, measured foliations, extremal length, fatgraphs, Schwarz-Christoffel mappings.

A key statement: any surface with at least one puncture is conformally equivalent to a surface obtained from half-infinite cylinder $[0,\infty) \times S^1$ by partitioning the boundary component into intervals and pairing them in a way so as to preserve arc-length.
There can be vertices of order 1, but these should correspond only to punctures; the infinite end of the cylinder corresponds to the desginated puncture. This representation is unique up to isomorphism. (This is a special case of a much more general structure theory for quadratic differentials and their relationship to measured foliations, which are a topological concept. Uniqueness is an elementary argument in the theory of extremal length).
The infinite cylinder is really $\log(w), where w ranges over a punctured disk, so the Poinare metric is immediate.

In your case: the boundary of the cylinder will fold into a Y, with its shape determined by the positions of three "folding" points on a circle whose pairwise distances must satisfy the triangle inequality. Given the glued-up cylinder, you can then solve for the uniformizing map to the sphere using the Schwarz-Christoffel technique generalizing the method of finding Riemann mappings to polygons; in general, you'd need to solve for the Schwarzian derivative. Then the problem is to match the crossratio of the four points
a,b,c,z to the four points that result. This may sound complicated, but it should
work numerically quite quickly once it's programmed: it's at least a finite-dimensional problem, and I think the solutions are numerically well-behaved if you do it right.

I suspect that for this particular case you can express the solution in terms of elliptic functions, but I'm not up to working out the formula.