Why does $\frac{X - aZ}{Z}$ have a double pole at the point $(0 : 1 : 0)$ and not just one (Divisors)? If I have an Elliptic Curve E and the function $\frac{X - aZ}{Z}$, I would have expected the divisor to be, defining a point $P = (a,b)$ and $-P = (a,-b)$, $div(f) = [P] + [-P] - [\infty]$.
Instead the correct solution would be $div(f) = [P] + [-P] - 2[\infty]$.
Where does this double pole come from? It would have made sense if the denominator would have been $Z^2$, but it is not.

A rational function $f(x) := x - a$, for example, expressed in homogeneous coordinates is $f(X/Z) = (X - aZ)/Z$ which has a single zero in the numerator and a single pole given by the denominator. Thus a simple zero and a simple pole always appear together. In general, when there are multiple zeros "up to multiplicity", there will be an equal number of poles "up to multiplicity".
For example, if $f(x) := (x-a)(x-b)$, when it is expressed in homogeneous coordinates, then it becomes $$f(X/Z) = (X - aZ)(X - bZ)/Z^2$$ which has two simple zeros in the numerator and one double pole given by the denominator. Thus, $div(f) = [a] + [b] - 2[\infty].$ The key idea is that both zeros and poles can have "multiplicity" and this has to be taken into account so that they balance each other in all cases. When two simple zeros merge they become one double zero, and so on.
