$$K^* = O_K^\times \times \pi^\Bbb{Z}$$
$$O_K^\times= \langle \zeta_{p^f-1}\rangle \times 1+\pi O_K$$
$$H = (1+\pi O_K)_{tors}$$
Since $H$ is torsion it contains some roots of unity, since they are equal to $1\bmod (\pi)$ it means they are $p^r$-th roots of unity, ie. $H = \langle \zeta_{p^a}\rangle$.
Because $\lim_{n\to \infty} (1+\pi x)^{p^n}=1$ then $(1+\pi x)^b$ is well-defined for $b\in \Bbb{Z}_p$ and $(1+\pi O_K)/H$ is a torsion free $\Bbb{Z}_p$-module, it is finitely generated because $1+p^2 O_K$ is of finite index in $1+\pi O_K$ and $\log$ is a $\Bbb{Z}_p$-module isomorphism $1+p^2 O_K\to p^2 O_K$,
that is to say $$(1+\pi O_K)/H= \prod_{j=1}^{[O_K:\Bbb{Z}_p]} (1+\pi x_j)^{\Bbb{Z}_p} H$$
$$1+\pi O_K = H \times \prod_{j=1}^{[O_K:\Bbb{Z}_p]} (1+\pi x_j)^{\Bbb{Z}_p} $$
$p^f$ is the size of $O_K/(\pi)$ and $\zeta_{p^a}$ is the largest $p^r$-root of unity in $K$.
I'll take a shot at answering.
$Q_0$: True, those are the definitions.
$Q_1$: Also true. To see a proof of the stated isomorphisms, look here Units of p-adic integers.
$Q_2$: Again, both isomorphisms are true. In the first one, you should be careful what you mean with exponentiation. $U^n$ means $\lbrace u^n\mid u\in U\rbrace$, which is a subgroup of $U$, but $(\Bbb{Z}_p/n\Bbb{Z}_p)^d$ means $d$ copies of $\Bbb{Z}_p/n\Bbb{Z}_p$, so stay away from mixing the two. The essence is that the former is multiplicative, and the latter is additive. Then I think you can convince yourself of the first isomorphism. The second isomorphism is not true in general, but uses the critical assumption that $(n,p)=1$, meaning that it has valuation $\nu_p(n)=0$, such that $n\Bbb{Z}_p=\Bbb{Z}_p$ as we discussed in your other question. Then $n(\Bbb{Z}_p^\Bbb{N})=(n\Bbb{Z}_p)^\Bbb{N} = \Bbb{Z}_p^\Bbb{N}$.
$Q_3$: The remainder of this question will be solved when you have a clear understanding of what $|\cdot|_\frak{p}$ is. Note that there is a choice to be made whenever you define the $\frak{p}$-adic norm, but it will always have the form $|\alpha|_\mathfrak{p} = p^{-c\nu_{\mathfrak{p}}(\alpha)}$ for some constant $c$. Neukirch has the convention of picking $c=[K:\Bbb{Q}_p]$, which he calls $d$ above. Also, it is clear that $\nu_\mathfrak{p} = e\nu_p$ where $e$ is the ramification index of the extension $K/\Bbb{Q}_p$.
Edit: I will adress your comments to this question.
You are correct in what Neukirch's convention is, it is equivalent to what I wrote. Now, $\nu_\mathfrak{p} = e\nu_p$ basically by definition: We have that $\mathfrak{p}^e = (p)$ in $\mathcal{O}_K$ by definition of $e$. If we let $\alpha\in \Bbb{Q}_p$, we have that $(\alpha)=(p)^{\nu_p(\alpha)}$ by definition, and then we see that
$$\mathfrak{p}^{\nu_\mathfrak{p}(\alpha)}=(\alpha)=(p)^{\nu_p(\alpha)} = (\mathfrak{p}^e)^{\nu_\mathfrak{p}(\alpha)} = \mathfrak{p}^{e\nu_p(\alpha)}.$$
I believe this holds in all characteristics. Hopefully this answers a couple of questions. (Edit 2) The leftmost equality is again the definition of $\nu_\mathfrak{p}$. The above of course implies $\nu_\mathfrak{p}(\alpha)=e\nu_p(\alpha)$ by unique prime ideal factorisation, which holds in any Dedekind domain, in particular any PID.
I do all of this using ideals as that is the neatest way of doing it, because you skip talking about units, but you can of course do this using usual prime element factorisation. If $\pi$ is any uniformizer, by definition $\pi^e=p\varepsilon$ for some unit $\varepsilon\in\mathcal{O}_K$. You can continue the rest of the story from there.
$|n|_\mathfrak{p}=1$ is true if and only if $n$ is a unit in $\mathcal{O}_K$, and for this it is sufficient to be a unit in $\Bbb{F}_p((t))$, and since $(n,p)=1$, $n$ is a unit even in $\Bbb{F}_p$. (Edit 2) $\nu_p$ is not a priori defined on $\mathcal{O}_K$, so the extension is literally by defining it to be $\nu_p=\frac{1}{e}\nu_\mathfrak{p}$. Remember that any ring homomorphism, in particular embeddings, preserve units.
(Third and final edit) I have rolled back your 8 edits to this answer. For transparency, your edits consisted of deleting some of my main points, adding references to Neukirch, adding other tiny details and giving an attempt to fill out the proof of $\nu_\mathfrak{p}=e\nu_p$ using elements.
These things are what I expect you to do for yourself. Do not ever replace what an answerer is saying with your own thoughts. If you are dissatisfied with my answer in its current form, wait for another answer, or ask a new question.
Best Answer
The reduction is a mapping of multiplicative groups. Think of the simplest case, with the $p$-adic integers: $\mathbf Z_p^\times \to (\mathbf Z_p/p\mathbf Z_p)^\times$ by $u \mapsto u \bmod p\mathbf Z_p$. The kernel is $1 + p\mathbf Z_p$, and $u \equiv v \bmod p\mathbf Z_p$ for units $u, v \in \mathbf Z_p^\times$ if and only if $u - v \in p\mathbf Z_p$, which can be written as a multiplicative coset relation: $u/v \in 1 + p\mathbf Z_p$. That would not be true if we weren't working with units.
If you don't see why the $(q-1)$-th roots of unity map bijective to the residue field, make sure to understand that in the simplest case of $\mathbf Z_p^\times$ first: why does $\mathbf Z_p^\times$ contain $(p-1)$-th roots of unity, and why do they reduce mod $p$ to the different elements of $(\mathbf Z_p/p\mathbf Z_p)^\times$? (Hint: Think of Hensel's lemma for $x^{p-1}-1$.)