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.
The calculation is straightforward. The capacitor size is simply a question of how much voltage drop you can tolerate over the duration of the pulse. The average current from the battery is a function of the duty cycle.
ΔV = I × Δt / C
Solving for C gives:
C = I × Δt / ΔV
Let's assume you can allow ΔV = 0.1V. For your first example, this works out to:
C = 25 mA × 25 ms / 0.1 V = 6.25 mF
The average current draw is 25 mA * 25 ms / 2.5 s = 0.25 mA.
For the second example, the numbers work out to:
C = 50 mA × 100 ms / 0.1 V = 50 mF
Average current = 50 mA * 100 ms / 1.0 s = 5 mA.
Best Answer
Good question. This is one way that primary (nonrechargeable) Lithium cells may differ dramatically from secondary (rechargeable) Li-ion cells so it is often a source of puzzlement. Namely, Li-ion (except LiFePO4) discharge and capacity curves typically gradually taper down from the initial 100% capacity voltage to the termination voltage. However the curve for Lithium coin cells is usually much less steep, i.e. lower slope or more flat, so the open circuit resting voltage near empty remains much closer to the initial 100% capacity voltage than it does in the Li-ion case.
For example, let's consider a typical discharge + resting voltage curve for a 3V LiMnO2 CR2032 coin cell (from here). The graph shows repeated discharge cycles, where each cycle consists of: first, a constant 1mA load for 11h, then 8.5hr rest, then 22mA pulse for 10 seconds, then a 30 minute rest. Notice that even when the coin cell is very close to empty (around 360h) the open-circuit resting voltage (2.7-2.8V) shown by the sawtooth peaks still remains very close to the initial 3.0V at 100% capacity.
Contrast that to an analogous typical Li-ion discharge curve below. Notice how the peaks drop off much more sharply towards the terminal voltage than they do above. Indeed the curve formed by the resting voltage peaks above is almost flat for most of the discharge, whereas the same curve below has a much greater downward slope.
The voltage rebounds after the load is removed because the cell has some nontrivial internal resistance \$R\$. This will cause a voltage sag of \$I\cdot R\$ at current \$I\$, which will disappear when the current drops back to 0. Such voltage rebound doesn't happen instantaneously due to time constants and exponential decay of various internal electrochemical processes (e.g. diffusion) that are components of the internal resistance. This is easier to see in the first graph where the sawtooth shape of the rebounding voltage is more evident. Symmetrically, there will also be \$I\cdot R\$ voltage downspikes when the current is increased which, is evident in the downward sawtooths in the first graph when the 1mA load is started, and also in the much larger downspikes at 22mA (but they don't last long enough to easily observe the sawtooth shape).
So, in summary, because primary Lithium coin cells have a flatter discharge curve than Li-ion cells, it is more difficult to use their open-circuit resting voltage to estimate remaining capacity. Testing them under load will yield better estimates, esp. if the load is nontrivial (e.g. notice how the curve formed by the lowest points of the high-current downspikes on the first graph tapers off much more sharply).