When i was doing the practice problems, I found the following problem: 
I was able to solve for everything except for V4.
When I looked at the solution manual,solution was as follows:
Use KVL: V4 = 3*2 + V3, (V3 was obtained from previous steps)
However, I'm confused because they wrote that it is a KVL equation, but it did not include 8 ohm resistor. I thought that in order to write a KVL equation, the entire loop must be completed.
I believe you're overcomplicating it. Since you already know $v_3$, all you need to do is calculate the voltage drop across the resistor with $3\Omega$. Since the current is $2A$, the voltage drop is $3*2$. Thus, $v_4$ should be 6 volts higher than $v_3$. This is a manifestation of KVL.
(more in depth if you are still confused)
At the core of KVL, it's that all around a loop the sum of the voltage drops must be 0. This is perhaps why you are thinking that you must look around a whole loop to get $v_4$. In truth, you already have without knowing it. Knowing $v_3$ means that you know the voltage drop from that point until "the end of your loop." The end of your loop can be defined wherever you'd like, but it most likely is defined to be at the positive terminal of the battery. Thus, you know the voltage all the way from the $v_3$ point of the circuit to the positive terminal of the battery. All that is left is the small portion of the loop with the $2\Omega$ resistor in it. After calculating that voltage drop, you have effectively calculated the sum of the voltage drops around the loop to be 0.
Another way of solving for $v_4$ if you know $v_3$ is to use KCL for node $v_4$ by summing the currents into the node
$2+ \dfrac{v_3-v_4}{3}=0$
On rearranging this gives the KVL equation
$v_3 + 3\times 2 - v_4 =0$
In terms of a potential diagram starting at the zero volt end of the 8 ohm resistor it looks like this:
