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.
If I understand you right, you're referring to the phenomenon seen in this picture (from the first Google hit), that near the horison the color of the sky is more light-blue (not exactly white):
Rayleigh scattering
The scattering in the atmosphere is for a large part Rayleigh scattering off of nitrogen and oxygen molecules, which are much smaller than the wavelength $\lambda$ of the light, and which has a preference for forward and backward scattering. If the Sun is more or less above you, the horison is roughly at 90º, which is the direction on which scattering is least likely.
Rayleigh scattering has an huge wavelength dependence of $\sim 1/\lambda^4$, meaning that blue light is scattered much more efficiently than red light. This is the reason for the blue sky in the first place, and the reason for the red sun at sunset.
Mie scattering
However, close to the ground — which is what you look at when you look at the horison — you have more water droplets which have a size of the order of the wavelength. In this case, the scattering is not Rayleigh scattering, but the more general solution of Mie scattering, which is not strongly wavelength dependent. Thus, all the colors are scattered more or less equally, producing a whitish glare.
(My first answer was not good, and did not take into account the Mie scattering. I'm sorry.)
Best Answer
Sure, let's do the order of magnitude calculation! According to Wikipedia:
The distance to Andromeda is $2.5 \times 10^6$ light years while the thickness of the Earth's atmosphere is on the order of $20$ kilometers. (It technically extends much higher, but the thickness decreases rapidly.) Then the ratio of the distances is $$\frac{\text{Andromeda distance}}{\text{atmosphere height}} \sim 10^{18}.$$ At this point we would naively multiple this by the ratio of densities and call it a day, but it's crucial to account for the loss mechanism.
Since only the hot regions matter, let's focus on those and suppose the whole line between the Earth and Andromeda is hot. Looking up standard numbers, for blue light we have $$N_2 \text{ Rayleigh cross section} \sim 2 \times 10^{-26} \, \text{cm}^2$$ and the Thomson cross section is wavelength independent, $$e^- \text{ Thomson cross section} \sim 7 \times 10^{-25} \, \text{cm}^2.$$ These are close enough that we can just neglect the difference, so we just need to compare the total distance and density. The ratio of the densities is about $10^{19}/10^{-3} = 10^{22}$, so we have $$\frac{\text{ISM effect}}{\text{atmosphere effect}} \sim \frac{10^{18}}{10^{22}} \sim 10^{-4}.$$ It looks like all the space between us and Andromeda has less effect than the atmosphere alone.
Edit: as pointed out by Joshua (a real astrophysicist, unlike me, a regular physicist who just multiplied a bunch of powers of ten), the ISR is much sparser outside of galaxies, so I should have used the size of a galaxy rather than the full distance between galaxies. Also, a much larger effect comes from Rayleigh scattering off interstellar dust, which comes out to 20%. This is relatively close to the effect of the atmosphere.