Do high power through hole resistors need a heat sink

The circuit as shown will work. Remember though that it's a linear regulator, which means that the voltage is dropped by turning excess energy into heat. At 4.5V the drop is small, but if you're dropping to 1.5V (3.5V drop) at 500mA (you shouldn't expect to draw more from USB) then you'll have to deal with 1.75W of heat. The amount of heat that the IC you linked can dissipate depends on the PCB design (did you mean to link a surface mount component?), but in any case 1.75W would be the upper end of what you could expect a TO-220 package component to dissipate. I'd probably use a heat sink or ensure that my load wasn't drawing as much current.

Resistor selection basically doesn't matter. The two aspects that you need to normally consider when selecting resistors are the power rating (1/4W, 1/2W etc) and the tolerance (1%, 5%, 10%). The power rating isn't important in this case (see below) and because you've got a manually adjustable potentiometer the tolerance really isn't important either. Almost any resistor of approximately the right value would do for \$R_2\$.

Both resistors can be low wattage ones. As you've said, the \$I_{ADJ}\$ current is negligible and can be completely ignored. There will also be a current flow through the two resistors from the output voltage to ground, which you can calculate with Ohm's law (\$V=IR\$). In all cases this will be about 7mA.

The capacitors are there for filtering/ripple reduction purposes and their characteristics aren't too important. If they're approximately the same as those suggested in the datasheet (and the voltage rating is above what they'll see) then there shouldn't be any issues.

Your calculations seem to correct. This is just a basic voltage divider calculation. The regulator adjusts the output voltage until it sees 1.24V on ADJ. You can confirm that your calculations are correct with the equation given on wikipedia:

$$V_{div} = V_{in} \times \frac{R_2}{R_1+R_2}$$

Where \$V_{div}\$ is 1.24V, \$V_{in}\$ is your target voltage and you want to solve for \$R_1\$.

$$R1 = \frac{V_{in} \times R_2 - V_{div} \times R_2}{V_{div}}$$

for the 4.5V case therefore:

$$R1 = \frac{4.5 \times 180 - 1.24 \times 180}{1.24} = 473 \Omega.$$

There's inherent variability in electronics which mean that whatever values you calculate won't be exactly right when you wire it up. Wire it up, connect the output to a volt meter and adjust the potentiometer until you have the right voltage.

First, you don't need a precision current source. The reason is that, at heart, a current source works by measuring the current through a resistor, then closes a feedback loop around the current measurement and a precision voltage source. Since you only want to measure the current (and therefor the charge) through your load, you don't need the precision reference voltage. You can use a fairly sloppy source, as long as you measure the current precisely.

So, how do you measure current? Well, that is (in principle) pretty straightforward. You just measure the voltage across a resistor in series with your load, generally called a shunt resistor. Of course, you have not indicated that extreme stability in the current level is required, and if it is you do need to worry about that.

Unfortunately, you've bitten off quite a lot with your requirements. You want quite a high current for your stability. This will play merry hell with your requirements, since self-heating will become a major player. Let's start with a baseline system. Let's figure you want 1 volt across your shunt at full current. Then the power dissipated will be 1/2 watt, and the target resistance will be 2 ohms. This will cause significant self-heating in the resistor. Go to digikey.com, and start looking at low-tempco resistors. Let's figure on using 10 ppm/deg C units. Restricting the search to in-stock resistors, you'll notice that the available higher-power units are still less than 1/2 watt, and they are generally not in stock, with minimum buys of 4000 units (admittedly, at 40 cents a pop, but that's still about 1600 bucks). Worse, they have high resistance values.

Once you get to 1/8 watt, you can find 10 ohm units. If you put 5 in parallel, you'll get 2 ohms at .625 watts. This, however, is going to be a no-go. The individual resistors are rated for a temperature of 70 C, or 50 degrees above ambient. This, of course will produce a thermal drift of nominally 500 ppm. In fact, if you were to find them, you'd need individual tempcos of about 0.2 ppm.

With this in mind, check out https://www.digikey.com/products/en/resistors/chip-resistor-surface-mount/52?k=&pkeyword=&pv2085=u10+Ohms&pv2=4&FV=ffe00034%2C4400c9&mnonly=0&ColumnSort=0&page=1&stock=1&quantity=0&ptm=0&fid=0&pageSize=25 and you'll find 0.2 ppm/deg, 10 ohm, 1/4 W. You can get 4 or 5 to put in parallel, and you should be OK. Not only is TCR low, PCR is 5 ppm at 70C. Granted, swallowing half of your error budget in a single source is just asking for trouble, but that's generally part of the game when doing things on the cheap. There is a very good reason why the current sources you've been looking at cost so much. Of course, they (the specified resistors) will run you about 50 - 60 bucks. Is that a problem? Is that "cheap"?

Well, it's certainly a lot cheaper than the sort of current source you've been looking at. And it's definitely a good idea to consider proper cooling for your shunt, but that will be a good idea anyways.

And while we're at it, you should be aware that your meter requirements are outside the usual boundaries of cheap. You require at least .001% linearity, and at least 5 1/2 digits from a DMM. If you're going to roll your own A/D, you need at least 17 bits.

And this sort of wide dynamic range and high accuracy imply sensitivity to input noise that you need to be aware of. Granted, if all you want to do is adding up the samples you'll get considerable averaging out of noise, although in this case a higher sample rate is better than low.

In neither case is it clear why you want such a low data acquisition rate. Sure, it's a lot of data, but unless you're going to have shifts of workers taking measurements, 200 hours of data is only 720,000 seconds. Assuming 10 bytes per sample, that's only a file size of 7.2 Mbyte. Even the lowly FAT32 can hold about 500 times that amount. On the other hand, even assuming 10 seconds per sample, are you really going to try to crunch 72 thousand data points by hand? for several setups? It is hard to imagine why it does not make sense to automate both the data acquisition and reduction. At the very least, you can do simple totalling in Excel almost trivially.

I'm inclined to agree with Tony Stewart that this is not a project for a beginner. If you absolutely must do it yourself, I'd go for a well-built shunt, and then go to with a commercial DAQ from a company like Measurement Computing. You can get an 8-channel, 24-bit DAQ with software that will do 2 samples/sec for a bit over 400 bucks. Input offset tempco is less than 0.5 uV/deg so you might not need to think about climate control for your instrumentation. Then again, gain tempco is on the order of 4 ppm/deg, so you probably do.

EDIT - Rather than use comments to reply to comments, I'm extending this answer.

I may, perhaps, have misunderstood your requirements. As I understand your post, you are interested in the total charge flowing through your load(s). You've said nothing about distinguishing between the current into the load and the current through the shunt resistor. In other words, you gave the impression that the load input equals the load output current, and if you measure the one you measure the other. Under these circumstances, there is little need for a precision source, at least not in the sense you seem to think it is. If you measure the current to 10 ppm, well, that's the best you can do. If it varies some between samples, then as long at that variation is not correlated with the sample interval it will all come out in the wash.

On the one hand, yes, stability at some level is necessary. My point was, however, that it doesn't need to be as great as you might think. Yes, if the current level changes with time it's necessary to track it. However, unless the changes (which can be considered noise) are correlated with the sampling time, long data runs will average out this noise. In other words, stability issues will tend to be filtered out over the long runs being considered. In principle, there is always the possibility that you can get accumulating errors, but this should not be much of a problem. And stability in this case means stability over 10's of seconds, which is not hard to do.

And I should quantify my terms, particularly stability. 0.01% (100 ppm) in a current source is not that hard or expensive, although 0.1% is much easier. And if you use the sort of low-tempco shunt I've suggested, you can use that voltage to control your current source, and the reference voltage becomes the limiting factor, followed by amplifier offset.

Additionally, temperature control is misleadingly easy to dismiss as "simple", and in some respects it is. However, unless you quantify your control, you have no way of knowing if it's adequate. You can't just supply a heat sink and be sure that the problem is solved. For that matter, you don't even know if there was a problem in the first place.