Consider the elliptic curve given by $E: Y^2 = X^3-X$ over the field $\overline{\mathbb{F}}_5$. I have computed the $\mathbb{F}_5$-rational points (in projective space, where $(0:1:0)$ is taken as the point at infinity): \begin{matrix} \textbf{1} & \textbf{2} & \textbf{3} & \textbf{4} \\ (0:1:0) & & & \\ (0:0:1) & (0:1:0) & & \\ (1:0:1) & (0:1:0) & & \\ (2:1:1) & (0:0:1) & (2:4:1) & (0:1:0) \\ (2:4:1) & (0:0:1) & (2:1:1) & (0:1:0) \\ (3:2:1) & (0:0:1) & (3:3:1) & (0:1:0) \\ (3:3:1) & (0:0:1) & (3:2:1) & (0:1:0) \\ (4:0:1) & (0:1:0) & & \\ \end{matrix} The numbers in the top row indicate the number of times I have added the element in the first column to itself. So each row below the top row is $a,2a, \dots ,na$, where $n$ is the order of $a$. I guess all the information in the table is not strictly related to the question.
Question: How can I construct a function $g \in k(E)$ that has a simple zero at each of the points in the table of order $4$, and a simple pole at each of the other points in the table, and no other zeros and poles on $E(\overline{\mathbb{F}}_5)$?
If I understand you aright, then try $$\frac{(x-2)(x-3)}y=\frac{x^2+1}{y}.$$ The numerator vanishes for $x\in\{2,3\}$, that is at the points of order $4$ and the denominator vanishes at the points of order $2$. On degree grounds you must have a pole at the point at infinity too.