Direction of current flow in Nodal Analysis

For starters, I don't understand how the numerator on these fractions ends up as anything but one....

I've left several comments pointing out careless errors in your equations but this deserves an answer even if the answer is elementary algebra.

Take, for example your first (corrected) equation:

$$\frac{1}{10}v_1 + \frac{1}{20}v_3 +\frac{1}{2}(v_3-v_2)+\frac{1}{2}(v_1-v_4)=0$$

Group the terms involving, for example, \$v_1\$

$$v_1(\frac{1}{10} + \frac{1}{2}) = v_1 \frac{3}{5}$$

Picture this:

^{simulate this circuit – Schematic created using CircuitLab}

It looks pretty obvious that \$V_a = V_1 + V_2\$, but this is technically a form of KVL. If you take a loop clockwise, you've got:

\$V_a - V_1 - V_2 = 0\$

which is the same thing that your intuition tells you.

Essentially, whenever you make one of those substitutions, you're making a loop and applying KVL around it. The only weird thing is that not all of the pieces of your loop need to be components in the circuit - it's perfectly valid to make one of them a voltage jump, like this example.

EDIT:

You asked for another circuit. Here's my best shot:

^{simulate this circuit}

If we go up the voltage supply and down the 'jump' in loop 1, we get \$V_a - V_1 = 0\$ or \$V_1 = V_a\$.

If we 'jump' to V1, move across R1, and 'jump' from V2 to ground, we get \$V_1 - V_{R1} - V_2 = 0\$ or \$V_{R1} = V_1 - V_2\$.

If we 'jump' to V2 and move back down R2, we get \$V_2 - V_{R2} = 0\$ or \$V_{R2} = V_2\$.

Hopefully those three examples make sense.