In the context of algebraic function fields, Abhyankar's Lemma states:
Let $F'/F$ be a finite separable extension of function fields. Suppose that $F'=F_1F_2$ is the compositum of two intermediate fields $F\subseteq F_1$, $F_2\subseteq F'$. Let $P'\in \mathbb{P}_{F'}$ be an extension of $P\in \mathbb{P}_{F}$ and set $P_i:=P'\cap F_i$ for $i=1,2$. Assume that at least one of the extensions $P_1|P$ or $P_2|P$ is tame. Then $$e(P'|P)=lcm\{e(P_1|P), e(P_2|P)\}$$
To keep this question short, I can provide desired definitions in the comments.
Now, my question is: what if we take two wild ramifications? Obviously, Abhyankar's Lemma does not hold. I found an example where it doesn't hold but I want to prove why it doesn't work. Unfortunately, I have no clue about the proof.
I would be grateful for any help.
I would build such an extension as a compositum of two Artin-Schreier extensions.
Let $F=\Bbb{F}_2(x)$ be the rational function field over $\Bbb{F}_2$. Consider the two quadratic extensions $$ F_1=F(y), y^2+y=x, $$ and $$ F_2=F(z), z^2+z=x^3. $$ Both $F_1/F$ and $F_2/F$ are Galois, the respective Galois groups generated by the order two automorphisms $y\mapsto y+1$ and $z\mapsto z+1$. Therefore the compositum $K=F_1F_2$ is also a Galois extension of $F$ with the Galois group isomorphic to the Klein four. The third intermediate field is $F_3=F(y+z)$, where $(y+z)^2+(y+z)=x^3+x$.
The standard tools (I use results from Stichtenoth's book, mine is the earlier edition from 1993) then imply easily that the infinite place $F_\infty$ of $F$ is totally (hence wildly) ramified in $F_i/F$ for all $i=1,2,3$. Therefore all the automorphisms of $Gal(K/F)$ belong to the inertia group. It follows that above $F_\infty$ there is a single place $K_\infty$ of $K$, and $e(K_\infty|F_\infty)=4$. Clearly $e(F_{i,\infty}|F_\infty)=2$ $F_{i,\infty}=K_\infty\cap F_i$, for all $i=1,2,3$.