15 W \$\times\$ 5 s = 75 J. If you want to store that in a capacitor you'll need a capacitance of
\$ C = \dfrac{2 \times 75 J}{(10 V)^2} = 1.5 F \$
There are supercaps with even higher capacitances, but types which can supply a 1.5 A current are between expensive and Damn Expensive™.
A battery will be a better choice. For instance a LiPo followed by a boost converter, to have a nicely regulated 10 V, even at varying battery voltage. Use another boost converter to charge the battery between load bursts; I would shut-down the charger while you're transmitting.
If the 3.3 V can supply 100 mA that's 330 mW. Suppose the two boost converters have an efficiency of 85 % then to get the required 75 J output you'll need 105 J input. At 330 mW that will take 315 seconds, that's 5 minutes and 15 seconds. So you can transmit a 5 second burst once every 5 minutes and 20 seconds. If you need a higher frequency you'll have to find another power source.
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I almost forgot: the battery has a less than 100 % efficiency too. You can't get every joule you put into it out again. So in practice the charging time will have to be longer than the 5 minutes.
I would use two LiPo cells in series. Then you'll have 7.4 V nominal, 8.4 V fully charged. That's relevant, because the closer the voltage to the 10 V the less current will be required from the battery. If you draw 1.5 A at 10 V a 3.7 V cell would have to supply 4.8 A if we take again the 85 % for the booster. Two cells in series will only have to supply 2.4 A.
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Dave's ultracap seems to be a good alternative, but he should mention that the 3.3 V to 2.5 V regulator also has to be a switching regulator, unless you're satisfied with a 75 % efficiency there. If we use eBay as a price reference you can buy a set of two 240 mAh LiPo's for 4.5 dollar, free shipping. That means they only will be discharged for 1 % during the 5 s burst. So while 14 dollar for the cap isn't bad, and definitely something to remember, the LiPo's are only a third of that.
It's also worth noting that stepping down from 3.3 V to 2.5 V and then up to 10 V is less efficient than up from 3.3 V to 7.4 V, and then to 10 V. That's because a switcher's efficiency is related to both the in-out voltage difference and their ratio.
Best Answer
What you are referring to is called a decoupling capacitor and is used to decouple the IC supply pins from the bus. In other words, it prevents a sensitive IC from being "starved" if another device on the bus turns on quickly and draws significant current, which would drop the bus voltage for a period of time. The capacitor supplies the extra current required to start up the device, as well as to prevent its chip from suffering the effects of a sudden loaded bus. This is generally required for high-speed devices that switch very quickly, as this tends to draw significant current. The capacitor is not necessarily chosen by its capacitance, but by its ESR (equivalent series resistance) and its ESL (equivalent series inductance). Ideally you would determine the speed at which the device would turn on, and pick the capacitor with the lowest ESR/ESL for that speed. The most common decoupling capacitor value is probably 0.1uF but for faster circuits you may require 0.01uF or 0.001uF (again, depending on their ESR and ESL at those speeds). If multiple devices with different speeds exist on the same bus, you may need more than one decoupling capacitor, one for each speed.
99 times out of 100 the datasheets will tell you exactly what value decoupling capacitors to use on which pins, so read the datasheet. This tutorial from Analog Devices is also a great resource.