Voltage Divider-like circuit with multiple sources
I can't quite determine the current flow because the lines from \$V_2\$ and \$V_1\$ carry current that is equal to \$V_2/R_2\$ and \$V_1/R_1\$ at the point in between the two resistors. How does this return to ground, and how would I begin to apply Kirchoff's Law for the loop?


Hmm. I don't agree with this. I think you came up with these equations with the old schematic, which had an extra ground connection.

The actual current through the resistors is
$$
I_1 = \frac{V_1 - V_{out}}{R_1}
$$
and
$$
I_2 = \frac{V_{out} - V_2}{R_2}
$$
Since there is no path for current to take another loop, these currents must be the same! That means that
$$
\frac{V_1 - V_{out}}{R_1} = \frac{V_{out} - V_2}{R_2}
$$
and some math gives you
$$
V_{out} 
= \frac{V_1 R_2 + V_2 R_1}{R_1 + R_2}
= \frac{V_1 R_2 + V_1 R_1 - V_1 R_1 + V_2 R_1}{R_1 + R_2}
= V_1 + (V_2 - V_1)\frac{R_1}{R_1 + R_2}
$$
Notice that if \$V_1 = 0\$, you get back the equation that you're used to:
$$
V_{out}|_{V_1 = 0} = V_2\frac{R_1}{R_1 + R_2}
$$