For a voltage divider circuit, why does \$V_{out} = I \times Z_{2}\$?
Because in \$V=R\cdot I\$ the \$R\$ is the resistor across which \$V\$ is measured. Since \$V_{out}\$ is usually measured against GND when the reference isn't marked, \$V_{out}\$ is the voltage across \$Z_2\$ in your example, and thus, the formula must be \$V_{out}=Z_2\cdot I\$.