I would like to understand how the Schwarz's lemma gives a bound for $|f'(z) - f'(a)|$ in the following theorem, which is a theorem of Conway's book:
$\textbf{1.4 Bloch's Theorem.}$ Let $f$ be an analytic function on a region containing the closure of the disk $D = \{ z \ ; \ |z| < 1 \}$ and satisfying $f(0) = 0$ and $f'(0) = 1$, then there is a disk $S \subset D$ on which $f$ is one-one and such that $f(S)$ contains a disk of radius $\frac{1}{72}$.
$\textit{Proof.}$ Let $K(r) = \max \{ |f'(z)| \ ; \ |z| = r \}$ and let $h(r) = (1 - r) K(r)$. It is easy to see that $h: [0,1] \longrightarrow \mathbb{R}$ is continuous, $h(0) = 1$, $h(1) = 0$. Let $r_0 = \sup \{ r ; h(r) = 1 \}$; then $h(r_0) = 1$, $r_0 < 1$ and $h(r) < 1$, if $r > r_0$. Let $a$ be chosen with $|a| = r_0$ and $|f'(a)| = K(r_0)$, then
$$ \textbf{(1.5)} \ |f'(a)| = (1 - r_0)^{-1}.$$
Now if $|z - a| = \frac{1}{2} (1 - r_0) = \rho_0$, $|z| < \frac{1}{2} (1 + r_0)$; since $r_0 < \frac{1}{2} (1 + r_0)$, the definition of $r_0$ gives
$\textbf{(1.6)} |f'(z)| \leq K(\frac{1}{2} (1 + r_0)) = h(\frac{1}{2} (1 + r_0)) [ 1 - \frac{1}{2} (1 + r_0) ]^{-1} < [ 1 - \frac{1}{2} (1 + r_0) ]^{-1} = \frac{1}{\rho_0}$
for $|z - a| < \rho_0$. Combining $(1.5)$ and $(1.6)$ gives
$$|f'(z) - f'(a)| \leq |f'(z)| + |f'(a)| < \frac{3}{2} \rho_0.$$
According to Schwarz's Lemma, this implies that
$$|f'(z) - f'(a)| < \frac{3 |z - a|}{2 \rho_0^2}$$
for $z \in B(a;\rho_0)$. Hence if $z \in S = B(a; \frac{\rho_0}{3})$,
$$|f'(z) - f'(a)| < \frac{1}{2\rho_0} = |f'(a)|$$
By lemma $1.4$, $f$ is one-one on $S$.
It remains to show that $f(S)$ contains a disk of radius $\frac{1}{72}$. For this define $g: B(0;\frac{\rho_0}{3}) \longrightarrow \mathbb{C}$ by $g(z) = f(z+a) - f(a)$, then $g(0) = 0$, $|g'(0)| = |f'(a)| = (2 \rho_0)^{-1}$. If $z \in B(0; \frac{\rho_0}{3})$, then the line segment $\gamma = [a,a+z]$ lies in $S \subset B(a;\rho_0)$. So by (1.6)
$$|g(z)| = |\int_{\gamma} f'(w) dw | \leq \frac{1}{\rho_0} |z| < \frac{1}{3}.$$
Applying lemma $1.2$ gives that
$$g(B(0;\frac{\rho_0}{3}) \supset B(0, \sigma)$$
where
$$\sigma = \frac{\left( \frac{\rho_0}{3} \right)^2 \left( \frac{1}{2\rho_0} \right)^2}{6 \left( \frac{1}{3} \right)} = \frac{1}{72}$$
If this is translated into a statement about $f$, it yields that
$$f(S) \supset B \left( f(a); \frac{1}{72} \right). \square$$
First of all, there is a typo when (1.5) and (1.6) are combined as it should be: $|f'(z) - f'(a)| \leq |f'(z)| + |f'(a)| < \frac{3}{2 \rho_0}$ as we can see from what comes above.
Second, an easy variant of Schwarz lemma says that if $|g(z)| \leq A$ on $|z| \leq R$, and $g(0)$=$0$ then $|g(z)| \leq \frac{A|z|}{R}$ there (just do the obvious change of variable $z$=$Ry$ and use the usual variant); by a simple translation same applies on a disc centered at some $a$, with now $g(a)$=$0$ and $|z-a|$ instead of $|z|$, so we get $|g(z)| \leq \frac{A|z-a|}{R}$ .
Putting this together the deduction follows with $g(z)$=$f'(z) - f'(a)$, $A$=$\frac{3}{2 \rho_0}$ and $R$=$\rho_0$