Kirchhoffs Law with 3 Voltage Sources

Label Vc and Vl in your diagram... current always flows from higher potential to lower potential (voltage). So depending on the direction your draw your current flowing, the voltage always drops in the direction of that current flow.

Simply put, in your first picture there is only one current loop. The current flowing through all the elements must be equal. I'll call the upper left corner node Vc and the upper right corner node Vl. Then namely:

(Vc - 0) / Zc = (Vl - Vc) / R = (0 - Vl) / Zl

Here Zc and Zl are the complex impedance of the capacitor and inductor respectively.

Alternatively, if we assume a current i0 is flowing through the circuit clockwise, and remember that voltage always drops across an element when current flows through it, then:

Vc = 0 - i0 * Zc
Vl = Vc - i0 * R
0 = Vl - i0 * Zl

Therefore:

0 = Vc - i0 * R - i0 * Zl = - i0 * Zc - i0 * R - i0 * Zl

Changing the sign:

i0 * Zc + i0 * R + i0 * Zl = 0

So yes, all elements should have the + sign.

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: $$ V_{out}|_{V_1 = 0} = V_2\frac{R_1}{R_1 + R_2} $$