[proofplan]
We construct explicit group homomorphisms for each isomorphism. For $U_K/U_K^{(1)} \cong k_K^\times$, we restrict the reduction map $\mathcal{O}_K \to k_K$ to the unit group. For $U_K^{(s)}/U_K^{(s+1)} \cong (k_K, +)$ with $s \geq 1$, we define a map $\phi_s: U_K^{(s)} \to k_K$ by $1 + \pi_K^s x \mapsto \bar{x}$ and verify it is a surjective group homomorphism with kernel $U_K^{(s+1)}$.
[/proofplan]
[step:Construct the isomorphism $U_K / U_K^{(1)} \cong k_K^\times$ via the reduction map]
Consider the surjective ring homomorphism
\begin{align*}
\rho: \mathcal{O}_K &\to k_K \\
x &\mapsto x \bmod \mathfrak{m}_K.
\end{align*}
Restricting to the unit group $\mathcal{O}_K^\times = U_K$: if $u \in U_K$, then $\rho(u) \neq 0$ (since $u \notin \mathfrak{m}_K$), so $\rho(u) \in k_K^\times$. Since $\rho$ is a ring homomorphism, the restriction
\begin{align*}
\rho|_{U_K}: U_K &\to k_K^\times
\end{align*}
is a group homomorphism (with respect to multiplication). It is surjective: for any $\bar{a} \in k_K^\times$, lift to $a \in \mathcal{O}_K$ with $\rho(a) = \bar{a}$; since $\bar{a} \neq 0$, we have $a \notin \mathfrak{m}_K$, so $a \in U_K$.
The kernel is $\ker(\rho|_{U_K}) = \{u \in U_K : u \equiv 1 \pmod{\mathfrak{m}_K}\}$. An element $u \in U_K$ satisfies $\rho(u) = 1$ iff $u - 1 \in \mathfrak{m}_K$ iff $v_K(u - 1) \geq 1$, which is the definition of $U_K^{(1)}$. By the first isomorphism theorem, $U_K / U_K^{(1)} \cong k_K^\times$.
[guided]
The reduction map $\rho: \mathcal{O}_K \to k_K = \mathcal{O}_K / \mathfrak{m}_K$ is the standard quotient homomorphism. We want to see what happens when we restrict it to the multiplicative group $U_K = \mathcal{O}_K^\times$.
An element $u \in \mathcal{O}_K$ is a unit iff $v_K(u) = 0$, iff $u \notin \mathfrak{m}_K$, iff $\rho(u) \neq 0$. So the restriction $\rho|_{U_K}$ maps $U_K$ into $k_K \setminus \{0\} = k_K^\times$. Since $\rho$ preserves multiplication ($\rho(uv) = \rho(u)\rho(v)$), this is a group homomorphism $U_K \to k_K^\times$.
Surjectivity: every nonzero element $\bar{a} \in k_K^\times$ has a lift $a \in \mathcal{O}_K$ with $\rho(a) = \bar{a}$. Since $\bar{a} \neq 0$, $a \notin \mathfrak{m}_K$, so $a \in U_K$.
Kernel: $\rho|_{U_K}(u) = 1$ means $u \equiv 1 \pmod{\mathfrak{m}_K}$, i.e., $v_K(u - 1) \geq 1$. This is precisely the condition $u \in U_K^{(1)} := \{u \in U_K : v_K(u - 1) \geq 1\}$. By the first isomorphism theorem, $U_K/U_K^{(1)} \cong k_K^\times$.
[/guided]
[/step]
[step:Define $\phi_s: U_K^{(s)} \to k_K$ and verify well-definedness for $s \geq 1$]
Fix $s \geq 1$. Every element $u \in U_K^{(s)}$ satisfies $v_K(u - 1) \geq s$, so $u - 1 = \pi_K^s a$ for some $a \in \mathcal{O}_K$, i.e., $u = 1 + \pi_K^s a$. Define
\begin{align*}
\phi_s: U_K^{(s)} &\to (k_K, +) \\
1 + \pi_K^s a &\mapsto \bar{a} := a \bmod \mathfrak{m}_K.
\end{align*}
We verify well-definedness: if $1 + \pi_K^s a = 1 + \pi_K^s a'$, then $\pi_K^s(a - a') = 0$, so $a = a'$ and $\bar{a} = \bar{a}'$. Thus $\phi_s$ is well-defined.
[/step]
[step:Verify $\phi_s$ is a group homomorphism]
Let $u_1 = 1 + \pi_K^s a$ and $u_2 = 1 + \pi_K^s b$ be elements of $U_K^{(s)}$. Their product is
\begin{align*}
u_1 u_2 = (1 + \pi_K^s a)(1 + \pi_K^s b) = 1 + \pi_K^s(a + b + \pi_K^s ab) = 1 + \pi_K^s c,
\end{align*}
where $c = a + b + \pi_K^s ab \in \mathcal{O}_K$. Since $s \geq 1$, $v_K(\pi_K^s ab) \geq s \geq 1$, so $\pi_K^s ab \in \mathfrak{m}_K$, and $\bar{c} = \bar{a} + \bar{b}$ in $k_K$. Therefore
\begin{align*}
\phi_s(u_1 u_2) = \bar{c} = \bar{a} + \bar{b} = \phi_s(u_1) + \phi_s(u_2).
\end{align*}
Thus $\phi_s$ is a group homomorphism from $(U_K^{(s)}, \cdot)$ to $(k_K, +)$.
[guided]
The crucial point is that the multiplicative structure of $U_K^{(s)}$ becomes additive modulo $\mathfrak{m}_K$. When we multiply $1 + \pi_K^s a$ and $1 + \pi_K^s b$, we get $1 + \pi_K^s(a + b) + \pi_K^{2s} ab$. The cross-term $\pi_K^{2s} ab$ has valuation $\geq 2s \geq s + 1$ (since $s \geq 1$), so upon reduction modulo $\mathfrak{m}_K$ it vanishes. This is why the multiplicative group $U_K^{(s)}$ "linearises" to the additive group $k_K$ at the level of successive quotients.
Concretely: $\phi_s(u_1 u_2) = \overline{a + b + \pi_K^s ab} = \bar{a} + \bar{b} + 0 = \phi_s(u_1) + \phi_s(u_2)$.
[/guided]
[/step]
[step:Identify the kernel as $U_K^{(s+1)}$ and conclude the isomorphism]
The kernel of $\phi_s$ is
\begin{align*}
\ker \phi_s = \{1 + \pi_K^s a : \bar{a} = 0\} = \{1 + \pi_K^s a : a \in \mathfrak{m}_K\} = \{1 + \pi_K^{s+1} b : b \in \mathcal{O}_K\} = U_K^{(s+1)}.
\end{align*}
The map $\phi_s$ is surjective: for any $\bar{a} \in k_K$, lift to $a \in \mathcal{O}_K$ with $\rho(a) = \bar{a}$, then $1 + \pi_K^s a \in U_K^{(s)}$ and $\phi_s(1 + \pi_K^s a) = \bar{a}$.
By the first isomorphism theorem,
\begin{align*}
U_K^{(s)} / U_K^{(s+1)} \cong (k_K, +).
\end{align*}
[/step]
