Why is the voltage divider formula used in this circuit to find the terminal voltage?
You are right that we can use Ohm's law. 

Look at the math: 

The current in this series circuit is: 

$$I = \frac{V_1}{R_1+R_2} $$ 

And the voltage drop across \$R_2\$ is equal to: 

$$V_{R_2} = I \times R_2 = \frac{V_1}{R_1+R_2} \times R_2 = V_1 \times \frac{R_2}{R_1 + R_2}$$  

$$V_{R_1} = I \times R_1 = \frac{V_1}{R_1+R_2} \times R_1 = V_1 \times \frac{R_1}{R_1 + R_2}$$  

As you can see the voltage divider formula comes directly from Ohm's law.