Despite what most textbooks claim, superposition of dependent sources is valid if done correctly.
There are three sources in this circuit so there will be three terms in the superposition.
For the first term, the two current sources are zeroed (opened) so \$V_x\$ is given by voltage division:
\$V_x = 10V \cdot \dfrac{4}{4 + 20} = \dfrac{5}{3}V\$
For the second term, the voltage source is zeroed (shorted), so the two resistors are now in parallel, and the 2A source is activated. Thus:
\$V_x = 2A \cdot 4\Omega || 20 \Omega = \dfrac{20}{3}V\$
Since the third, dependent source is in parallel with the 2A source, the last term has the same form:
\$V_x = 0.1 V_x \cdot 4\Omega || 20 \Omega = \dfrac{1}{3}V_x\$
Now, it's crucial at this point to not try and solve the previous equation (you'll only get \$V_x = 0\$ if you do.)
Rather, proceed with the superposition sum and then solve.
\$V_x = \dfrac{5}{3}V + \dfrac{20}{3}V + \dfrac{1}{3}V_x\$
Grouping terms:
\$V_x (1 - \frac{1}{3}) = \dfrac{25}{3}V\$
Solving:
\$V_x = 12.5V\$
KVL states that the sum of voltages around a circuit (loop) is zero. Multiple interconnected loops create a mesh. This allows you to write down a system of equations that can be solved for all of the voltages (and currents) in the mesh. So yes, KVL and mesh analysis are closely related.
Similarly, KCL states that the sum of currents into (or out of) a node is zero. This allows you to write down a different system of equations that can be solve for all of the currents (and voltages) in the circuit (collection of interconnected nodes). KCL and nodal analysis are closely related.
Superposition is an approach used to deal with linear circuits that have multiple indpendence sources, in which you evaluate each independent source, one at a time, and then add the results together. Mesh or nodal analysis might be used for the individual solutions.
Best Answer
It's the 1 ohm resistance of r2. As other answers point out, this step is basic current division. Using conductances, \$G = \frac{1}{R}\$, current division has the form of voltage division (in fact, current division is the dual of voltage division).
Voltage division for two series connected resistors:
\$V_{R_1} = V_S \dfrac{R_1}{R_1 + R_2}\$
Current division for two parallel connected resistors:
\$I_{R_1} = I_S \dfrac{G_1}{G_1 + G_2} \$
Note that these equations are duals. If you know one of these, you get the other by duality.
Applying the 2nd equation to your problem:
\$I_2 = I_1 \dfrac{G}{G + g_2} = I_1 \dfrac{\frac{1}{4}}{\frac{1}{4} + \frac{1}{1}} = I_1 \dfrac{1}{1 + 4}\$