The point is that the family $\{k_n : n \in \mathbb{N}\}$ is uniformly bounded, hence normal. Let's start with an $f\in S$. We define $f_1 \colon z \mapsto \frac{f(z)}{w_f}$. By definition of $w_f$, we have $\mathbb{D} \subset f_1(\mathbb{D})$ and $1 \notin f_1(\mathbb{D})$. So with $f_2 = f_1 - 1$, $U := f_2(\mathbb{D})$ is a simply connected domain containing $D_1(-1)$, and not containing $0$. So there are two branches of the square root defined on $U$, we choose the one with $\sqrt{-1} = i$ and define $f_3\colon z \mapsto \sqrt{f_2(z)}$. Since $D_1(-1) \subset f_2(\mathbb{D})$, we have $\sqrt{D_1(-1)} \subset f_3(\mathbb{D})$, and an easy computation shows $D_{\sqrt{2}-1}(i) \subset \sqrt{D_1(-1)}$.
Since a branch of the square root on a domain $V$ can never attain both values, $w$ and $-w$ (if we had $\sqrt{a} = w$ and $\sqrt{b} = -w$, then $a = (\sqrt{a})^2 = w^2 = (-w)^2 = (\sqrt{b})^2 = b$, so $w = -w$ and hence $w = a = 0$, but a domain on which a branch of the square root exists cannot contain $0$), we have $D_{\sqrt{2}-1}(-i) \cap f_3(\mathbb{D}) = \varnothing$. Then for $f_4 = f_3 + i$ we have $\lvert f_4(z)\rvert \geqslant \sqrt{2}-1$ and finally for $f_5 = \frac{1}{f_4}$ (which corresponds to $k_n$) we obtain $\lvert f_5(z)\rvert \leqslant \frac{1}{\sqrt{2}-1} = \sqrt{2} + 1$ for all $z\in \mathbb{D}$. So
$$S_5 = \biggl\{ \frac{1}{i + \sqrt{f/w_f - 1}} : f \in S\biggr\}$$
is a uniformly bounded, hence normal family. Now we trace back the construction and see that normality is preserved. For that, we consider a locally uniformly convergent sequence $(k_n)_{n\in \mathbb{N}}$ in $S_5$ and deduce that the corresponding sequence in $S$ has a locally uniformly convergent subsequence. The only step that needs a nontrivial argument is the first step, going from $(k_n)$ to $(1/k_n)$. We denote the limit function by $k$. Since all $k_n$ never attain the value $0$, we either have $k \equiv 0$, or $k$ never attains the value $0$ (Hurwitz). But $k_n(0) = \frac{1}{2i}$ for all $n$, hence $k(0) = \frac{1}{2i}$, and so $k$ never attains the value $0$. In particular, for every $r\in (0,1)$ there is a $\delta_r > 0$ such that $\lvert k_n(z)\rvert \geqslant \delta_r$ for all $z$ with $\lvert z\rvert \leqslant r$ and all $n$. Thus $\frac{1}{k_n} \to \frac{1}{k}$ uniformly on $\overline{D_r(0)}$. Since $r$ was arbitrary, and every compact subset of $\mathbb{D}$ is contained in some $D_r(0)$, we have the compact (or equivalently locally uniform) convergence of $\frac{1}{k_n}$ to $\frac{1}{k}$. Subtracting the constant $i$ from every function of course preserves locally uniform convergence, and since $z \mapsto z^2$ is continuous, we also have locally uniform convergence
$$\frac{f_n}{w_{f_n}} - 1 = \biggl(\frac{1}{k_n} - i\biggr)^2 \to \biggl(\frac{1}{k} - i\biggr)^2,$$
and adding the constant $1$ to all functions of course doesn't change that. So we know that the sequence $\frac{f_n}{w_{f_n}}$ is locally uniformly convergent. If we knew that the sequence $(w_{f_n})$ converges, it would follow that the full sequence $(f_n)$ converges compactly. But since we don't know that, we extract a subsequence for which $w_{f_{n_m}}$ converges, and deduce compact convergence for the corresponding subsequence $\bigl(f_{n_m}\bigr)_{m\in \mathbb{N}}$.
This finishes $(iii)$, and in the argument for normality of $S$, we have also shown the normality of $S' = \{g_f : f \in S\}$, that is, the first part of $(iv)$.
For the second part of $(iv)$, let $r_0 = \inf \{ \lvert w_f\rvert : f \in S\}$. Choose a sequence $(f_n)$ in $S$ with $\lvert w_{f_n}\rvert \to r_0$. Since for every $f\in S$ the function $f_{\varphi} \colon z \mapsto e^{-i\varphi} f(e^{i\varphi}z)$ also belongs to $S$, we can assume that $w_{f_n} = \lvert w_{f_n}\rvert > 0$ for all $n$. By normality, we can assume that the sequence is locally uniformly convergent, say to $g$. Then we have $g(0) = 0$ and $g'(0) = 1$, so $g$ is not constant, and by Hurwitz' theorem, $g$ is also a schlicht function, i.e. $g\in S$. Then for $0 < r < \lvert w_g\rvert$, and all large enough $n$ we have $\overline{D_r(0)} \subset f_n(\mathbb{D})$, by Rouché's theorem, so $r < r_0$. It follows that $r_0 = \lvert w_g\rvert > 0$.
Incidentally, we have $r_0 = \frac{1}{4}$, but the proof of that is a bit involved and technical. The value $\lvert w_f\rvert =\frac{1}{4}$ is attained by the Koebe functions
$$f_{\alpha} \colon z \mapsto \frac{z}{(1 - \alpha z)^2},$$
where $\lvert\alpha\rvert = 1$.
