For the jump from equation 1 to 2 you can follow this procedure
\begin{align}
2\exp(-(t/9)^2)
&\geq \mathbb{P}\left (\sqrt{\frac{d}{cnL^2}}\sum_{i=1}^n (f(x_i) - \mathbb{E}[f])z_i \geq t \right ) & (\text{Eq. 1})\\
=& \mathbb{P}\left (\sqrt{\frac{cL^2}{dn}} \sqrt{\frac{d}{cnL^2}}\sum_{i=1}^n (f(x_i) - \mathbb{E}[f])z_i \geq \sqrt{\frac{cL^2}{dn}}t \right ) \\
=& \mathbb{P}\left (\frac{1}{n} \sum_{i=1}^n (f(x_i) - \mathbb{E}[f])z_i \geq \sqrt{\frac{cL^2}{dn}}t \right )
\end{align}
where the second line is just multiplying both sides of the inequality by the same positive constant, and the third is just simplifying.
Now choose $t = \sqrt{\frac{dn}{cL^2}}\frac{\epsilon}{8}$. Plugging this into the first and last gives
\begin{align}
\mathbb{P}\left (\frac{1}{n} \sum_{i=1}^n (f(x_i) - \mathbb{E}[f])z_i \geq \sqrt{\frac{cL^2}{dn}} \sqrt{\frac{dn}{cL^2}}\frac{\epsilon}{8} \right ) &\leq 2\exp\left (-\left (\frac{1}{9} \sqrt{\frac{dn}{cL^2}}\frac{\epsilon}{8}\right )^2 \right ) \\
\mathbb{P}\left (\frac{1}{n} \sum_{i=1}^n (f(x_i) - \mathbb{E}[f])z_i \geq \frac{\epsilon}{8} \right ) & \leq 2\exp\left (-\left (\frac{1}{9} \sqrt{\frac{dn}{cL^2}}\frac{\epsilon}{8}\right )^2 \right ) \\
&=2\exp\left(-\frac{1}{9^2}\frac{\epsilon^2}{8^2}\frac{dn}{cL^2}\right) \\
&\leq 2\exp\left (-\frac{\epsilon^2nd}{9^4cL^2} \right)
\end{align}
which proves the first part.
For the second part of the question, we need the basic fact that if "not $B$" implies "not $A$," then $\mathbb{P}(A) \leq \mathbb{P}(B)$. To see this, note that
\begin{align}\mathbb{P}(A) &= \mathbb{P}(A \text{ and } B) + \mathbb{P}(A \text{ and not } B) \\
&=\mathbb{P}(A|B)\mathbb{P}(B) + \mathbb{P}(A|\text{not } B)\mathbb{P}(\text{not } B) \\
&\leq 1\mathbb{P}(B) + 0\mathbb{P}(\text{not } B) = \mathbb{P(B)}
\end{align}
With this in mind, let $B$ be the event $\left| \frac{1}{n}\sum_{i=1}^n z_i \right | \geq \frac{\epsilon}{8}$ and let $A$ be the event
$$\exists f \in \mathcal{F} \text{ s.t. } \frac{1}{n}\sum_{i=1}^n \mathbb{E}[f]z_i \geq \frac{\epsilon}{8}.$$
Now note that since $|\mathbb{E}[f]| \in [0,1]$, we have if not $B$ then
\begin{equation}
\frac{\epsilon}{8} > \left| \frac{1}{n}\sum_{i=1}^n z_i \right | \geq |\mathbb{E}[f]|\left| \frac{1}{n}\sum_{i=1}^n z_i \right | \geq \frac{1}{n}\sum_{i=1}^n \mathbb{E}[f]z_i
\end{equation}
which mean not $B$ implies not $A$, and we can apply the bound above.
