To turn off VS2, replace it with a short circuit. The circuit then consists of R1 in series with the parallel combination of R2 and R3.

To turn off $V_{S2}$, replace it with a short circuit. The circuit then consists of $R_1$ in series with the parallel combination of $R_2$ and $R_3$.
The equivalent resistance of $R_2$ and $R_3$ in parallel is:

$$R_{23} = \frac{R_2 \times R_3}{R_2 + R_3} = \frac{470\,\Omega \times 470\,\Omega}{470\,\Omega + 470\,\Omega} = \frac{220900\,\Omega^2}{940\,\Omega} = 235\,\Omega$$

The total resistance seen by $V_{S1}$ is:

$$R_{total1} = R_1 + R_{23} = 470\,\Omega + 235\,\Omega = 705\,\Omega$$

The total current flowing from $V_{S1}$ is:

$$I_{total1} = \frac{V_{S1}}{R_{total1}} = \frac{12\,V}{705\,\Omega}$$

This current splits between $R_2$ and $R_3$. Since $R_2 = R_3$, the current divides equally. The current through $R_2$ due to $V_{S1}$ ($I_{R2,S1}$) is:

$$I_{R2,S1} = \frac{I_{total1}}{2} = \frac{12\,V / 705\,\Omega}{2} = \frac{6\,V}{705\,\Omega} \approx 0.008511\,A = 8.511\,mA$$

The direction of this current is downwards through $R_2$.
The voltage across $R_2$ due to $V_{S1}$ ($V_{R2,S1}$) is:

$$V_{R2,S1} = I_{R2,S1} \times R_2 = \frac{6\,V}{705\,\Omega} \times 470\,\Omega = \frac{2820}{705}\,V = 4\,V$$

The polarity is positive at the top of $R_2$.

To turn off $V_{S1}$, replace it with a short circuit. The circuit then consists of $R_3$ in series with the parallel combination of $R_1$ and $R_2$.
The equivalent resistance of $R_1$ and $R_2$ in parallel is:

$$R_{12} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{470\,\Omega \times 470\,\Omega}{470\,\Omega + 470\,\Omega} = \frac{220900\,\Omega^2}{940\,\Omega} = 235\,\Omega$$

The total resistance seen by $V_{S2}$ is:

$$R_{total2} = R_3 + R_{12} = 470\,\Omega + 235\,\Omega = 705\,\Omega$$

The total current flowing from $V_{S2}$ is:

$$I_{total2} = \frac{V_{S2}}{R_{total2}} = \frac{18\,V}{705\,\Omega}$$

This current splits between $R_1$ and $R_2$. Since $R_1 = R_2$, the current divides equally. The current through $R_2$ due to $V_{S2}$ ($I_{R2,S2}$) is:

$$I_{R2,S2} = \frac{I_{total2}}{2} = \frac{18\,V / 705\,\Omega}{2} = \frac{9\,V}{705\,\Omega} \approx 0.012766\,A = 12.766\,mA$$

The direction of this current is downwards through $R_2$.
The voltage across $R_2$ due to $V_{S2}$ ($V_{R2,S2}$) is:

$$V_{R2,S2} = I_{R2,S2} \times R_2 = \frac{9\,V}{705\,\Omega} \times 470\,\Omega = \frac{4230}{705}\,V = 6\,V$$

The polarity is positive at the top of $R_2$.

Since both $I_{R2,S1}$ and $I_{R2,S2}$ flow in the same direction (downwards through $R_2$), they add up.

$$I_{R2} = I_{R2,S1} + I_{R2,S2} = \frac{6\,V}{705\,\Omega} + \frac{9\,V}{705\,\Omega} = \frac{15\,V}{705\,\Omega}$$