Let $L/K$ be a local field extension.
Question: Is it always true that $\operatorname{Gal}(L^{nr}/K) \simeq \hat{\mathbb{Z}} \times I$ (where $L^{nr}$ is the maximal unramified subextension of $\bar{L}/L$ and $I$ be the inertia group of $\operatorname{Gal}(L/K)$)?
In Example 5 of Tim and Vladimir Dokchitser's Euler Factors determine local Weil Representation, we have the case $K = \mathbb{Q}_{13}$, $L=K(\sqrt[3]{13})$ and $I = C_4$.
In case this is true, it would be really helpful to understand this result (for instance, explaining where $\hat{\mathbb{Z}}$ and $I$ come from). As I still struggle a lot with Algebraic Number Theory and Galois Theory (for example not knowing basic results regarding Frobenius elements, maximal unramified extensions, infitnite Galois Theory in general etc.) it would be the best if you could do it in basic terms.
Thank you in advance!
Almost everything in your question is incorrect. First of all please stop with "local field", you mean $K$ is a finite extension of $\Bbb{Q}_p$. Then $$K^{nr}=\bigcup_{p\ \nmid\ n}K(\zeta_n)$$
$\phi\in Gal(K^{nr}/K)$ is the Frobenius sending $\zeta_n $ to $\zeta_n^q$ where $q= |O_K/(\pi_K)|$.
$\pi_K=\pi_{K^{nr}}$ and the reduction $\bmod (\pi_K)$ of $\phi$ is the Frobenius of the residue field. The Galois group of $Gal(K^{nr}/K)$ is the profinite completion of $\langle \phi \rangle$ which is $$\varprojlim\langle \phi\rangle / \langle \phi^m\rangle\cong \varprojlim \Bbb{Z/mZ}\cong \widehat{\Bbb{Z}}$$
Proof: $Gal(L^{nr}/K^{nr})\to I(L/K)$ is injective and its number of elements is the ramification degree $e(L/K)=e(L^{nr}/K^{nr})$.
$$Gal(L^{nr}/K)=Gal(L^{nr}/(L^{nr})^\Phi)\rtimes Gal(L^{nr}/K^{nr})$$
In general $\rtimes $ is not a direct product.