Thèvenin equivalent circuit
You're not combining series and parallel resistors correctly. To use the voltage divider equation with your second circuit, you first need to combine \$R\$ and \$R_2\$ in parallel with \$2R\$:

$$\frac{1}{R_{parallel}} = \frac 1 {2R} + \frac 1 {R + R_2}$$

$$\frac 1 {R_{parallel}} = \frac 1 {2R} + \frac 1 {R + \infty} = \frac 1 {2R} + 0 = \frac 1 {2R}$$

$$R_{parallel} = 2R$$

The voltage between the \$2R\$ resistors is then:

$$V_{mid} = V_1 \frac{2R}{2R + 2R} = \frac {V_1} 2$$

which is what you got for the first circuit.

I'm not sure why you didn't include the \$R\$ resistor in your Thevenin resistance. Normally in this kind of problem you want to end up with only one equivalent resistor.