Your question is valid and is very good you are thinking this way so young.
First, error propagation is a whole area of its own and there is not a unique way to do so, nor there is an absolute best way to be provided. So much so that the ISO document used for having a consensus on this, is called "Guide to the Expression of Uncertainty in Measurement", not a manual.
This is the case mainly because determining uncertainties is very much a compromise (depends on how much confident you want to have on your value) and model guided (depends on your assumptions of what you don't know).
What does this means?
Well first it is impossible to have absolute confidence on a value, so when you say a rod's length $20.0\pm0.2cm$ (and please note the number of significant digits should match) you are assuming that the value can be anything between $19.8cm$ and $20.2cm$ but there is no chance it will be lower nor higher.
Secondly, you are assuming that all values between $19.8cm$ and $20.2cm$ are equally likely, which is not usually the case.
Now don't get me wrong: your reasoning is not flawed, is completely legitimate, although not the most practical, and therefore not what is generally used. All this was to clarify what it really means your error propagation and your expectations of them falling inside a range, etc.
How we do things
From experience we have seen that many values, when measured exhibit a Normal distribution, which means in few words, that repeated measurements on the same conditions will be concentrated around the mean value, and the further from this value the less you find. Since this is very often, we commonly assume that this is the case in general. So in this sense you are correct in expecting that your rod's value $20cm$ should be in the center of the interval.
As for the interval to report, when assuming normal distribution, it is common practice to report the interval whose center is the mean value of the measurements, and which will contain roughly 63% of all possible values. The logical reasons for this are numerous and lengthy to cover here, but still this is a convention.
Now you see the information behind a value: $20.0\pm0.2cm$ means generally that we assume the Normal distribution for the values, we estimated $20.0cm$ to be the mean, and 63% of the measurements will be between $19.8cm$ and $20.2cm$.
What does it all have to do with error propagation?
Well if it all depends on our definition of a value and its uncertainty, then it has everything to do.
To propagate errors we don't just sum $L_1$ and $L_2$ max and min values, and this is precisely because we know that these values are not 100% inside their intervals, they are 63% inside. And so the values of $L_t = L_1 + L_2$ will not be 63% inside the interval between $L_1min+L_2min$ and $L_1max+L_2max$. This is not easy to prove in concise manner with your math skills, but you seem curious so you will find it out.
So the way we propagate errors in this case is $\delta L_t = \sqrt{(\delta L_1)^2 + (\delta L_2)^2}$ and although its advantages can be argued logically, this is still based on the Normal distributed model, and the 63% interval requirement.
So what about the area?
Well as you saw above, your idea of propagating by using the minimum and maximum values changes the coverage of your interval. Above, when a sum is involved, it will not in general yield a 63%. Here the same happens, you can not expect to have a correct $63%$ interval because, for example $W_{min}*L_{min}$ gives a value of the area that actually is outside of the 63% interval for the area. The prove of this involves probability and again you will have to take my word here.
The formula used is in fact $\delta A / A = \sqrt{(\delta W / W)^2 + (\delta L / L)^2}$ which needs some basic calculus knowledge to deduce, but I will try to give some sense to it from your point of view. See how similar it is to the previous one? Is like the same as before, only you are using $\delta value/value$ which you can think of as a relative uncertainty: is how important is the uncertainty in relation to the value it characterizes.
Indeed, if you would like to compare the uncertainty between your height, say $700\pm2mm$ and say the Earth's radius $696342\pm65km$. Now which one is measured more precisely? It would be hard to say here since $2mm$ is much smaller than $65km$, but the values are very different too. However looking at the relative uncertainties you will see that Earth's value is indeed more precise!
Finally, let me say that although I only spoke of the general case encountered in physics, it is not always correct to assume the Normal distribution, nor even expect that the mean is in the center of the interval. Under some conditions you may encounter Poisson distribution in which the mean is usually closer to the smallest values of the integral. Also in other cases, particularly in research, we expect to have values much more rigorously measured, and intervals are chosen to contain as much as 95% of the measured values, and sometimes more.
