separable Galois group of $\mathbb{F}_q((T))$ Let $K$ be the field $\mathbb{F}_q((T))$, where $q = p^r$ for some prime $p$.
What is the structure of $G_K := {\rm Gal}(K^{sep}/K)$?
I believe the maximal unramified extension is $\overline{\mathbb{F}_q}((T))$, and after that surely all you can do is adjoin prime to $p$-th roots of $T$. Thus, I believe we have the exact sequence
$$1\rightarrow G_{\overline{\mathbb{F}_q}((T))}\rightarrow G_{\mathbb{F}_q((T))}\rightarrow {\rm Gal}(\overline{\mathbb{F}_q}((T))/\mathbb{F}_q((T)))\rightarrow  1$$
Here, $I_K := G_{\overline{\mathbb{F}_q}((T))}$ is the inertia group, which I think is just the prime-to-$p$ completion of $\mathbb{Z}$? (in this case there isn't any wild inertia right?)
Also, the last piece in the exact sequence is just $G_{\mathbb{F}_q}\cong\widehat{\mathbb{Z}}$. So, the middle piece $G_{\mathbb{F}_q((T))}$ is a metabelian group. Is it abelian? What is it's abelianization? What is the image of $I_K$ in its abelianization? What is the action of $G_{\mathbb{F}_q}$ on $I_K$?
 A: I realized I should probably just put my comments as an answer. It's somewhat difficult since the most immediate answer to your question is 'no', because it's not true that there is no wild ramification. So, I mention the answers to some of your other questions and interesting side notes.
To save myself a lot of parenthesis-ing, let $K=\mathbb{F}_q((T))$. 
Your analysis is flawed because $K$ does, in fact, have wildly ramified extensions. For example, any Eisenstein polynomial of degree $p$ for $\mathcal{O}_K$ gives you such an example (so, perhaps, you could take $f(X)=X^p-T^2X-T$ and adjoint a root). In fact, the wild ramification is the only 'hard part' of $G_K$ since, as you correctly observed
$$\text{Gal}(K^\text{tame}/K)\cong \widehat{\mathbb{Z}}^{(p)}\rtimes \widehat{\mathbb{Z}}$$
where one can explicitly describe the action as follows. Namely, 
$$K^\text{tame}=K^\text{ur}(\{T^{\frac{1}{n}}:(n,p)=1\})$$
and so you can think of the elements of $\widehat{\mathbb{Z}}^{(p)}=\text{Gal}(K^\text{tame}/K^\text{ur})$ as being (systems) of automorphisms of the form $T^{\frac{1}{n}}\mapsto \zeta T^{\frac{1}{n}}$ with $\zeta$ an $n^\text{th}$ root of unity. It's then clear how
$$\widehat{\mathbb{Z}}=G_{\mathbb{F}_q}=\text{Gal}(K^\text{ur}/K)=\text{Gal}(K(\{\zeta_n:(n,p)=1\})/K)$$
acts.
So, to answer your other questions.

Is it abelian?

No. Explicitly, write down an $S_3$-extension $K$ of $\mathbb{F}_q(T)$ totally ramified at $T$. Then, $K_\mathfrak{p}/\mathbb{F}_q((T))$ (where $K_\mathfrak{p}$ is the completion of $K$ at the prime above $T$) is an $S_3$-extension of $\mathbb{F}_q((T))$.

What is its abelianization?

This is the content of local class field theory for completed function fields. Namely, that gives us an isomorphism 
$$G_K^\text{ab}\cong \widehat{K^\times}$$
where, here, one must be careful that the hat does not denote profinite completion but, instead, completion with respect to the topology given by abelian norm subgroups (subgroups of the form $N_{L/K}(L^\times)$ with $L/K$ fintie abelian)--these two notions agree in the mixed characteristic case, but not in the equicharacteristic case. In particular, you can use this to write down infinitely many abelian totally ramified (and thus wildly ramified) cyclic $p$-extensions of $\mathbb{F}_q((T))$. 
NB: I assumed that you were asking about the Hasudorff abelianization (i.e. the abelianization in the category of Hausdorff topological groups: $G^\text{ab}=G/\overline{[G,G]}$). The actual abelianiation is, as far as I know (which is not very far), untenable--such a quotient has no 'Galois theoretic meaning' since you're quotienting by a non-closed subgroup in general.

What is the image of $I_K$?

Under local class field theory $I_K^\text{ab}$ gets carried to $\widehat{\mathbb{F}_p[[T]]^\times}$ with the same caveat about the meaning of the hat.

What is the action of $G_{\mathbb{F}_q}$ on $I_K$?

I answered this above.
Bonus:
Your analysis is correct for studying $G_{k((T))}$ if $k$ is an algebraically closed field of characteristic $0$. Namely, there we have the two equalities:
$$k((T))^\text{tame}=k((T))^\text{sep},\qquad k((T))^\text{ur}=k((T))$$
and thus, using your analysis, we get that 
$$G_{k((T))}\cong \widehat{\mathbb{Z}}$$
with the covers explicitly being given by $k((T))(T^{\frac{1}{n}})$.
This makes intuitive sense because, intuitively, we imagine that $k((T))$ is the fucntions on a very small ('formally small') punctured disk around $0$ in $\mathbb{A}^1_k$. So, for example, if $k=\mathbb{C}$ the fundamental group of such a punctured disk is $\mathbb{Z}$, and so the etale fundamental group (which is the Galois group of $k((T))$) is its profinite completion: $\widehat{\mathbb{Z}}$.
Hope it helps!
