Lets say we have a wierstrauss normal form elliptic curve $C : y^2 = x^3 + Ax^2 + B$. Then we look at the vertical line in the function field $f/g = (x - a)/1 \in K(C)$ which intersects at points $(a, b)$ and $(a, -b)$.
Lets say I want to calculate the poles of $f$, so homogenizing wrt $Z$, we get $F = Z f(X/Z, Y/Z) = Z\left(\frac{X - aZ}{Z}\right)$ and $G = Zg(X/Z, Y/Z) = Z$ so $$ F/G = \frac{X - aZ}{Z} \in K(C)_{\textrm{proj}}$$ is our rational projective function. We then immediately see the pole is when $Z = 0$. This means there's a triple pole at $(0 : 1 : 0)$.
But if instead we homogenize wrt $Y$, we instead get $F = Yf(X/Y, 1) = Y\left(\frac{X - aY}{Y}\right)$ and likewise $G = Y \implies$ $$ F/G = \frac{X - aY}{Y} $$ and so the pole instead is when $Y = 0$ which are the points $(e_i : 0 : 1)$ where $i = 1, \dots, 3$ and $x^3 + Ax^2 + B = (x - e_1)(x - e_2)(x - e_3)$. But these poles are completely different to what we got before.
What is wrong in my calculations above and how do we reconcile this? Thanks
The expression $F/G = \frac{X-aZ}{Z}$ is correct, but you are not using it correctly. You can't simply look at the numerator and denominator to find poles, because (for example) "$Z$" by itself is not a well-defined function on $\mathbb{P}^2$. (I mean, imagine we had a function $h([X:Y:Z]) = Z$. Then $[0:0:1] = [0:0:2]$ on $\mathbb{P}^2$, but $h([0:0:1]) = 1$ and $h([0:0:2]) = 2$, which is nonsensical -- functions need to have equal values on equal inputs.) Even if you want to use $\frac{X-aZ}{Z}$ to build some naive intuition, you need to do so very carefully. For example your "pole" $[0:1:0]$ is also a zero of the numerator of $\frac{X-aZ}{Z}$, so you need to calculate very carefully the order of vanishing of the numerator as well as the denominator in order to determine the order of vanishing of this pole. (And in fact the order of vanishing of this pole is 2, not 3 as you claim.)
The expression $F/G = \frac{X-aY}{Y}$ is not correct. Indeed, if it were correct, then by transitivity of equality we would have $\frac{X-aZ}{Z} = \frac{X-aY}{Y}$, which implies $\frac{X}{Z} = \frac{X}{Y}$; this is obviously not true on all of $\mathbb{P}^2$, as we can easily see by taking (for example) $X=1, Y=2, Z=1$.
If you want to see this calculation done correctly, see this answer.