I have not done the math but would expect that the radiation from the asphalt as $T^4$ will favor larger gradients for higher temperatures. I have the impression that air goes something like $T^6$, so even conduction energy transferred will have larger gradients the hotter it is. Your g is temperature dependent I guess.
Edit in response to edit of question.
Shouldn't that happen on cold days as well as on hot days?
I am copying from the comments below:
if you do the calculation of black body, asphalt at $40^\circ \mathrm{C}$ radiates $547\mathrm{watts}/\mathrm{m}^2$. At $41^\circ\mathrm{C}$ 551, i.e. difference $4\mathrm{watts}/\mathrm{m}^2$, whereas at $50^\circ\mathrm{C}$ 617 and at $51^\circ\mathrm{C}$ 628, i.e. $11\mathrm{Watts}/\mathrm{m}^2 \Delta T$. that is what I mean by g is temperature dependent. If there is convection the asphalt remains cooler and the equilibrium between heating from incoming radiation and cooling from black body is lower.
On a cold day even if windless with no convection the equilibrium temperature will be much lower since it starts heating from a much lower ground temperature. You can fry eggs in the summer in Greece on the asphalt, not in the winter.
In addition $5^\circ\mathrm{C}$ per meter is a low number. The asphalt may be at $50^\circ\mathrm{C}$ or $60^\circ \mathrm{C}$ but at 1 meter it will not be more than 35, in Greece, summer. In northern latitudes maybe $25^\circ\mathrm{C}$?
1) Yes, in the real world, only a tiny portion of the light scatters by the Rayleigh scattering. This may be reinterpreted as the simple fact that generic places in the blue sky are far less bright than the Sun. It means that the generic places of the sky become blue but the Sun itself remains white. For the same reason, distant mountains keep their color. Also, the distant mountains don't increase the amount of blue light from other directions much simply because the intensity of light reflected from distant mountains into our eyes is vastly smaller than the intensity of light coming directly (or just with Rayleigh scattering) from the Sun to our eyes. And even if it were not smaller, e.g. when the Sun is right below the horizon and the mountains are needed, we won't be able to easily distinguish that the blue sky actually depends on the mountains.
Rayleigh scattering is caused by particles much smaller than the wavelength, i.e. individual atoms and molecules, so it doesn't really matter which of them they are. The rate of Rayleigh scattering is therefore more or less proportional to the air density which means that a vast majority of it occurs in the troposphere, especially the part closer to the surface.
2) The changing atmospheric density only impacts the angle of the propagation of the sunlight substantially if the atmospheric density changes at distance scales comparable to the wavelength. If the length scale at which the density changes is much longer than that, the impact on the direction of light is negligible and calculable by Snell's law.
If you ever watch Formula 1 races, you may see some fuzzy waving water-like illusion near the hot asphalt. This is indeed caused by density fluctuations caused by the variable heat near the asphalt (well, a campfire could have been enough instead of Formula 1). However, in this case the direction of light only changes slightly because the regions of hot and cold air are still much longer than the wavelength (half a micron or so).
If you think about ways how to get density fluctuations comparable to the wavelength which is really short, you will see that the source is in statistical physics and the naturally fluctuating air density due to statistical physics is actually nothing else than an equivalent macroscopic description of the Rayleigh scattering! When you calculate the Rayleigh scattering, you may either add the effect of individual air molecules; or you may directly calculate with a distribution of many air molecules and the source of the effect is that their density isn't really constant but fluctuates. So these two calculations are really equivalent. They are the microscopic and macroscopic description of the same thing, like statistical physics and thermodynamics.
If the blue light manages to come from a direction that differs from the direction of the source of light, the Sun, then – assuming that the atmosphere doesn't emit blue light by itself, and it doesn't (at least not a detectable amount of it) – it is scattering by definition. To get a substantial change of the direction, you need small particles, and that's by definition Rayleigh scattering. So there's no other source of the blue sky than the Rayleigh scattering – although the Rayleigh scattering may be described in several ways (microscopic, macroscopic etc.).
Well, there's also the Mie scattering – from particles much larger than the wavelength, especially spherical ones, like water droplets. However, for the Mie scattering to be substantial, you need a substantial change of the index of refraction $n$ inside the spheres, which is OK for water. Also, the Mie scattering is much less frequency-dependent (because $n$ only slightly depends on the frequency, nothing like the fourth power here) than the Rayleigh scattering so it doesn't influence the overall color much. Not only during the sunset, some grey-vs-white strips on the clouds near the horizon are caused by the Mie scattering. The Rayleigh scattering really has a monopoly on the substantial change of the color.
Best Answer
Yes -- because refraction influences the apparent distance to the horizon, it also has an effect on the curvature.
To visualize this, it might help to think in extreme cases, for example in the case where due to refraction the horizon is at an apparent distance of only 1 meter. In this case, the curvature of the horizon would be extreme (it would be a circle of radius 1 meter around you).
In reality the curvature effect is much smaller of course, and I doubt that it is visible.
(As an aside: one result of atmospheric refraction that is observable is a phenomena called 'the green flash'. Because refraction is colour-dependent, the red and yellow part of the sun could already have set, while a small part of the 'green sun' is still above the horizon. This can be observed by the naked eye, preferably when the horizon is sharp.)