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.
The answer to which one you use depends on the application, and the efficiency needs.
For example, you're asked to make a phone charging dock. The dock is powered via a 12 V wallwart, and powers the phone with 5V of power at 500mA. Using a linear regulator, 3.5W is dissipated.That's quite a bit of waste, but you're connected to the mains, and a charging dock is a big enough device, where a properly heat sunk regulator wouldn't cause a lot of heating issues.
On the flip side, suppose you're building a wearable device that operates on a small Li-Po battery, even if you designed a LDO circuit that only wastes about 1W of power, a switching circuit would be more desirable as if designed properly, you could reduce your wastage to <10% that of the linear regulator
Note: Pay attention to the efficiency curves of switching regulators. They normally only have high efficiency for small ranges of current usage, and it helps to understand what current usage your application operates on in different condition to design the most efficient power circuit. Also - laying out swtiching regulators on a PCB can be hit/miss - I've seen a lot of incidents where tiny layout issues can mess with the desired voltage out.
Best Answer
First things first - a SMPS is not a simple circuit, and they just get more and more sophisticated. So one option you have is to start with a toy implementation with an inductor, a BJT, a capacitor and a microcontroller to drive the BJT. You wont get 2A out of it on your first go, and you'll probably fry a few BJTs, but you'll learn a lot.
But if you're keen on building something useful, then you need to pick an IC. It's hard to find much simpler than TI's Simple Switcher series. The LMZ14202H would probably meet your needs.
Both the inductor and the switcher are integrated, so you only have to concentrate on the feedback circuit. Follow their application notes very carefully and you should have something running fairly quickly. The LMZ14202H only comes in an SMD package though, so you'll probably want to get a 0.05" breakout board too for prototyping.
Trying to re-use components from existing power supplies is unlikely to be a fruitful path - those components are likely to be quite specialised, chosen to suit a specific design, and reverse-engineering a commercial power supply design is harder than building one from scratch.