Electrical – Applying KVL on RC circuit to get the natural reponse v_c(t>0)

Strictly speaking, it's not the sign of the resistors that can be changed, it's the sign (direction) of the current \$I_x\$. Resistors always have positive resistance but you can choose the current in either direction as long as you are consistent.

Suppose you chose \$I_x\$ in the opposite direction. Starting from node \$f\$ and moving counterclockwise, you have a positive voltage across the \$6\Omega\$ resistor, another positive voltage across the \$3\Omega\$ resistor, and a positive voltage across the 12V supply. This gives:

\$I_{x}(6\Omega + 3\Omega) + 12 = 0\$

which results in \$I_{x} = -4/3\$A. The sign is different, indicating that the current actually flows in the opposite direction (i.e. in the original direction shown in the image).

The reason your 4A answer is wrong is that you weren't consistent with the \$I_x\$ direction for the two resistors -- you had \$I_x\$ flowing in the opposite direction through the \$3\Omega\$ resistor.

Look at the first equation you have shown: $$2*I_c + \frac{1}{4} \frac{dV_c}{dt} - y(t)=0$$ The first term is Voltage, the second term is Current, the third is Voltage. So the second term cannot be correct, it should simply be the voltage across the capacitor. $$2*I_c + V_c - y(t)=0$$ There are three dependent variables and you only have two equations. So you need one more equation/relationship. Add the equation that relates current to voltage of a capacitor: $$I_c =\frac{1}{4} \frac{dV_c}{dt}$$ Now you can take combine the three equations into one equation with a single dependent variable y(t) as requested by your problem statement.