This is not correct. If it was the compositum then you would get that the absolute Galois group is a product of pro-$p$ group and pro-$p$' groups (i.e. inverse limit of prime to $p$ finite groups). This is obviously wrong, since for example, if you take prime $\ell$ such that $\ell=1\pmod p$. Then the nonabelian group $C_\ell \rtimes C_p$ (with respect to an embedding of $C_{p}\to Aut(C_{\ell})$) is realizable over $\overline{\mathbb{F}}_q((t))$ but is not a product of $p$-group and prime-to-$p$ group.

BTW: $C_m$ denotes a cyclic group of order $m$

This is not correct. If it was the compositum then you would get that the absolute Galois group is a product of pro-$p$ group and pro-$p$' groups (i.e. inverse limit of prime to $p$ finite groups). This is obviously wrong, since for example, if you take prime $\ell$ such that $p=1\pmod \ell$. Then the nonabelian group $C_p \rtimes C_\ell$ (with respect to an embedding of $C_{\ell}\to Aut(C_{p})$) is realizable over $\overline{\mathbb{F}}_q((t))$ but is not a product of $p$-group and prime-to-$p$ group.

BTW: $C_m$ denotes a cyclic group of order $m$