This appears to be an unconventional way to report correlation (or lack thereof). It focuses more on the variability of the measurements (across the earth at each fixed altitude) than on the correlation among them. As such the graphic may be of physical interest but it's an obscure way (at best) of comparing two measurement systems.
At each vertical position (an estimated altitude based on a pressure reading) the plots summarize between 3 and 18 pairs of data obtained over fixed stations on the earth's surface. The summaries consist of sample standard deviations normalized by the LIDAR readings (the reference measurement).
When comparing measurements, one is usually interested in assessing their correlations. We need to do a little math to relate this graphic to those correlations. Let $(X,Y)$ be a random variable representing the (LIDAR, GOMOS) readings. Let the variance of $X$ be $\sigma^2$, the variance of $Y$ be $\tau^2$, and their correlation equal $\rho$. Then
$$Var(X-Y) = Var(X) + Var(Y) - 2Covar(X,Y) = \sigma^2 + \tau^2 - 2 \rho \sigma \tau.$$
Consequently we can recover the correlation from the covariances:
$$\rho = \frac{1}{2\sigma \tau}(Var(X) + Var(Y) - Var(X-Y)).$$
Let the LIDAR mean be $m$. The plots depicts estimates of $\sigma/m$ (relative LIDAR SD): call this $s$; $\tau/m$ (relative GOMOS SD): call this $t$; and $\sqrt{Var(X-Y)}/m$ (relative SD of difference): call this $r$. Plug the estimates in to the preceding formula:
$$\rho = \frac{1}{2(s m)(t m)}((s m)^2 + (t m)^2 - (r m)^2;$$
$$\rho = \frac{s^2 + t^2 - r^2}{2 s t}.$$
These are, of course, estimates of $\rho$, subject to sampling uncertainty.
We can now qualitatively identify several portions of the plot:
$s = t = r$, approximately, between 35 and 45 km. From the formula we estimate $\rho \sim 1/2$. This is modest correlation--not very good for two measurements of the same thing.
One of $s$ and $t$ is small relative to the other and $r$ is comparable to the larger. This occurs from 20 to about 25 km and 45 to 50 km. The formula indicates $\rho \sim 0$. This is lack of correlation.
$r$ is small and $s$ and $t$ are comparable (between 28 and 35 km). Now we estimate $\rho \sim 1$. This is what one hopes to see for two consistently comparable measurements.
In short, good correlation occurs when the green line lies substantially to the left of the red or blue lines and there is lack of correlation wherever the green line approximates (or exceeds) either or both of the red and blue lines. Overall, correlation is poor except between 27 and 34 km.
It sounds like you're talking about what's sometimes called a regressogram, with a log-scaled x-variable.
There are a number of issues here, not necessarily in logical order:
the quantity you're plotting is a mean, so if you want to plot median absolute deviation, it's the MAD of the means you want.
your suggestion $\text{MAD}/\sqrt n$ leads to the question "when is the MAD of the mean equal to the MAD of the data divided by $\sqrt n$?"
when you say "it seems that median absolute deviation is a better estimator than mean absolute deviation" ... that depends what we're talking about - a better estimator of what?, and under what circumstances?
So, "when is the MAD of the mean equal to the MAD of the data divided by $\sqrt n$?"
The answer is, unlike the situation with standard deviation, this is not generally the case. The reason why standard deviations of averages scale as they do is that variances of independent random variables add (more precisely, the variance of the sum is the sum of the variances when the variables are independent), irrespective of the distributions of the components (as long as the variances all exist). It is this particular property that largely accounts for the popularity of variances and standard deviations.
Neither the median deviation, nor the mean deviation have that property in general.
However, when the data are normal, they will in effect inherit that property, since the ratio of the population mean deviation or median deviation to the standard deviation at a normal will be a constant, normals are closed under convolution, and standard deviations scale that way.
If the data were reasonably close to normal, it could perhaps be adequate.
What else might be done? One way to estimate the standard error of a statistic is via the bootstrap; for the mean deviation - being a mean - this should do well in large samples. Unfortunately, medians don't do so well under the bootstrap, and this issue will carry over to median absolute deviations.
If you have some probability model for your data, there's also simulation as a way of approaching the problem.
Best Answer
Probably a line for the mean and a line for +/- twice the standard deviation. That would be the "default" plot for that.
That said, I think you may be missing the point of the plot with the single line for the standard deviation. If what you're trying to represent is change in the variability of Y over X (i.e., heteroscedasticity), then a line plotting SD over X might work. It really does depend on the data and the questions that you're trying to ask. There just isn't a set of rules that you can follow to produce good plots every time, and in general the more automated the plotting system gets, the more useless I find it.