Shunt capacitor as discontinuity in transmission line
So, how is \$ V_C (t) = V^+(1 - \exp{(-t/\tau_c)}) \$ obtained? 


Your schematic model is correct. The left transmission line can be modeled with a Thevenin equivalent as you showed and the right transmission line can be modeled with just a resistor equivalent to its characteristic impedance.

You seem to understand why the right-hand model works, so I'll focus on the left one.

First, why \$Z_0\$ in series? Imagine that there was actually a wave propagating from the right. Then the model of the left-side transmission line would have to look (to that wave) just like you've done for the model on the right (by superposition,  you can ignore any sources on the left when working out how the circuit responds to this hypothetical signal arriving from the right). So you'd have an equivalent input impedance (looking in from the right) of \$Z_0\$. The \$Z_0\$ in your schematic represents this.

Second, why the source? Imagine there was no capacitor and you were just looking for the signals (\$V\$ and \$I\$) at an arbitrary point in an infinite transmission line. You'd be able to work out the behavior of the waveforms. To make your Thevenin equivalent, you just need to choose a voltage source that would give the same behavior. You can continue to use this same model after adding the capacitor, because the presence of the capacitor doesn't change the model of the portion of the transmission line to the left of it.

Note: you should actually have \$ V_C (t) = \frac{V^+}{2}(1 - \exp{(-t/\tau_c)}) \$


  And why is \$ \tau_c = CZ_0 / 2 \$ and not \$ \tau_c = CZ_0 \$?


You have \$Z_0/2\$ instead of just \$Z_0\$ because your schematic model is correct and both "resistors" (actually equivalent circuits of transmission lines) are in parallel with the capacitor.