Discharging RLC. Kirchhoff's second law
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.