Voltage drop across a single resistor and across two resistors
In the first circuit, you have one (single) voltage source and one (single) resistor.

This one (single) resistor is connected directly across the voltage source terminals ( terminals \$A\$ and \$B\$). Thus, from point \$B\$ to \$\$A the voltage is equal to the battery terminal voltage \$V_B\$ and because our single resistor is also connected directly between these two-point (B and A), The resistor must &quot;see&quot; the same voltage across his terminals as is &quot;given&quot; by the battery. And this is why \$V_B = V_1\$.
The voltage across the battery equal the voltage across the resistor.
But for the second case, we have a different situation.

Again we have one (single) voltage source but this time we have two resistors connected in series. And again the voltage across terminals \$A\$ and \$B\$ is equal to the battery voltage. But now neither of resistors is connected directly across the battery terminal voltage. So the voltage drop across the resistors will split because our two resistors are connected in series thus in a series circuit, the current that flows through each of the components is the same (only one path for current to flow).
\$V_B = V_1 + V_2 = IR_1 + IR_2\$
How can i calculate Vs in this circuit knowing Vo=2?
And some water analogy example of a series circuit.

And some water analogy for parallel connection. Notice that this time all resistor will see the same voltage (VB) but the current will split betwen resistors.