How do I calculate the equivalent RLC circuit values for a non-ideal capacitor?
The contribution of the 22uF ideal capacitor at 5MHz is practically nothing, as you can confirm from Xc  which is around 1m\$\Omega\$. 

The step of +/-20mV is due to the resistance, which must be about 50\$\Omega\$ * 0.02V/10V = 0.1\$\Omega\$.

The extra 200mV of voltage at the end of the ramp (at which time the signal generator voltage has reached its maximum amplitude) is entirely due to the inductance. The current rate of change is approximately 400mA/10ns so we know the inductance L = V/(di/dt) must be ~5nH. 

Below is an LTspice .asc file if you want to play with it. 

Version 4
SHEET 1 880 680
WIRE 416 0 208 0
WIRE 64 112 -64 112
WIRE 208 112 208 0
WIRE 208 112 144 112
WIRE -64 160 -64 112
WIRE 416 208 416 160
WIRE 432 208 416 208
FLAG -64 240 0
FLAG 432 272 0
SYMBOL res 160 96 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R1
SYMATTR Value 50
SYMBOL voltage -64 144 R0
WINDOW 3 -72 195 Left 2
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V1
SYMATTR Value PULSE(-10 10 0.1ns 10ns 10ns 100ns 200ns)
SYMBOL res 400 64 R0
SYMATTR InstName R2
SYMATTR Value 0.1
SYMBOL cap 416 208 R0
SYMATTR InstName C1
SYMATTR Value 22µ
SYMBOL ind 400 -16 R0
SYMATTR InstName L1
SYMATTR Value 5n
TEXT -232 240 Left 2 !.tran 0 1.2u 0.6u 0.1n uic