You are right in that a switcher makes a lot more sense for your application (12V in, 5V 1.5A out) than a linear regulator. A linear would waste 7V * 1.5A = 10.5W in heat, which would be challenge to get rid of. For linear regulators, current in = current out + operating current. For switchers power in = power out / efficiency.
I haven't looked up the TI part you mention (I might have if you had supplied a link). There are two broad classes of switching regulators, those with internal switches and those that drive external switches. If this regulator is the second kind, then dissipation in the part won't be a problem since it's not handling the power directly.
If it is a fully integrated solution, then you do have to look at dissipation. You can compute this dissipation from the output power and the efficiency. The output will be 5V * 1.5A = 7.5W. If the switcher is 80% efficient, for example, then the total input power will be 7.5W / 0.8 = 9.4W. The difference between the output power and the input power is the heating power, which in this case is 1.9W. That's way better than what a linear regulator would do, but is still enough heat to require some thought and planning.
80% was just a number I picked as a example. You need to look at the datasheet carefully and get a good idea what efficiency is likely to be at your operating point. Good switcher chips have lots of graphs and other information about this.
Once you know how many Watts will be heating the chip, you look at its thermal spec to see what the temperature drop from the die to the case is. The datasheet should give you a degC per Watt value. Multiply that by the Watts dissipation, and thats how much hotter the die will be than the outside of the case. Sometimes they tell you the thermal resistance from the die to ambient air. This is usually the case when the part is not intended to be used with a heat sink. Either way, you find how many deg C hotter the die will be than anything you can cool or deal with.
Now you look at the max die temp, then subtract off the above temp drop value. If that's not at least a little above your worst case ambient air temperature, then you have a problem. If so, it gets messy. You either need a heat sink, forced air, or use a different part. Higher power switchers are usually designed for external switch elements because power transistors come in cases intended to be heat sunk. Switcher chips usually don't.
I don't want to go on speculating, so come back with numbers about your particular situation, and we can continue from there.
IMO the two main question would be :
1) What microcontroller do you use? You might be able to use it at a lower voltage, maybe by decreasing the clock frequency. This would simplify your problem a gread deal.
For example if you want to use an ATMega, you may power it with as low as 1.8V if you keep the clock frequency at or below 4MHz. It will also be happy at 8MHz with much less than 3.3V. Then you can power it straight from the battery/power manager chip; not using a regulator at all will give you some significant battery gains.
2) What is the 3V3 part's lowest rated voltage? Depending on that, and the possible gains you'd make by not regulating the mcu voltage, you may be able to just discard the last drops of juice from your lipo for the same battery life (there is not that much energy left when it drops below 3.3V anyway... plus, that'd be healthier for the battery), and get rid of the step-up to 3.3V altogether. Now a LDO regulator might to the job. No switching regulator means less noise ; less parts, less power usage, less volume/weight and less price also might mean a possible bigger battery...
In order to make the right choices you'd need to specify the specifications of the parts you'd use. But using an unregulated voltage for most of the circuit, including MCU, and regulated voltage for only the 3.3V parts has many advantages.
Best Answer
A switching power supply will be noisier than a linear regulator, no question. The only way to know if it will disturb the wifi is to power it up and see what happens.
Any switching regulator will dissipate power as a function of conduction losses and switching losses. These parts with an integrated MOSFET do need PCB cooling - usually a multilayer PCB with relatively large 'islands' of copper interconnected with vias. Again, you'll have to do some math to figure out a loss estimate, and gauge the copper size accordingly.
The datasheet says that the 5V part at 2A load and 12V in operates at around 83% efficiency. So, for 10W out you're losing just over 2W in the device (and the external diode that completes the buck converter). This will scale down somewhat with your reduced loading (conduction losses will drop, switching losses probably won't).