Not getting readings from capacitor

There are two questions: the number of charging and the time of charge.

The 2450 battery has about 620mAh of energy stored. (dataheet from digikey). I'm assuming that you want to charge the capacitor with the same voltage as the battery: 3v. Therefore to fully charge a 470uF you'd need:

\$ Q_{cap} = C \times U \$

\$ Q_{cap} = 470 \mu F \times 3V = 1.41 mC \$

\$ Q_{bat} = 620mAh \$

\$ Q_{bat} = 620mAh \times \dfrac{3600 s}{1 h} \times \dfrac{1 A}{1000 mA} = 2232 As = 2232 C \$

\$ Cycles = \dfrac{Q_{bat}}{Q_{cap}} ~= 1582978 \$

Therefore the battery can charge the capacitor over 1.5 million times. If you want to "fully charge" the capacitor with 45v this number would drop to 100 thousand times, more than enough.

Remenber that I'm not assuming losses or other circuitry attached to the battery

The second question is if the capacitor can be charged under 6 seconds. We need to know how is the max currency we can get from the battery. Back to the datasheet this number is 0.2mA.

\$ Q_{cap} = 1.41 mC \$ (from earlier calculations)

In order to feed this amount of charges in 6 seconds we would have a currency of:

\$ I_{cap} = \dfrac{Q_{cap}}{\Delta T} = \dfrac{1.41mC}{6 s} = 0.235mA \$

As the required currency is lower than the one needed, there is no way to charge the capacitor under 6 seconds. The number from the datatheet is the "continuous standard load". It may be possible to charge the capacitor in less than 6 seconds, but it may be dangerous.

You may try another manufacturer aside the one I showed, they may have a similar battery with more "continuous standard load".

Another point to mention is that you need to consider the current drawn from the circuit you're building to charge the capacitor and subtract it from the available current from the datasheet.

What's the purpose of the diode? If that's an LED, then it's clamping the voltage at around 1.9V, hiding most of the exponential curve that you are looking for.

In other words, the rising edge of your waveform is just the first part of an exponential curve that is asymptotically headed toward 5V, but gets stopped at 1.9V by the clamping action of the diode. The falling edge is the entire view of an exponential curve that starts at the diode clamp voltage, and is headed toward 0V — but gets interrupted by the next charge cycle.