EDIT: This question discusses why the sky is blue with regards to classical rayleigh scattering.
Light does indeed travel in straight lines. However, this just indicates the direction of the movement of the wave as a whole. A light wave is an electromagnetic oscillation, and therefore higher frequencies will have quicker oscillations. The wave itself will be moving forward at the same speed, but there will be "more waviness" per unit length for a higher frequency wave than for a lower frequency wave.
If you switch from the so called "wave picture" to the so called "particle picture" (the shabbily named wave-particle duality) then the frequency $\nu$ of a photon is related to it's energy as given by $$ E = h \nu$$
These photons have an intrinsic position-momentum uncertainty wherein their position at any given moment cannot be described to an arbitrarily small accuracy. It is the probability of finding the photon at a given place that you have referred to as the quantum amplitude of a photon.
The square of the quantum amplitude describes the probability of the photon (neglecting time dependence for the moment) of being in a particular place in space, or possessing a particular momentum etc., depending on what basis you are describing the system in. So for example, if $\psi(x)$ is a function describing the quantum amplitude of a particle at every point in space, the probability that the particle lies somewhere between $0$ and $x$ is given by $$ P(x) = \int_{0}^{x} |\psi(x)|^2 dx $$
One cannot predict it's position more definitively than this probability.
Now that I've clarified what I mean by a quantum amplitude - There does exist scattering theory for quantum mechanics. This scattering theory deals with the problem of how these quantum amplitudes change when confronted by an external "potential". A potential can also be a potential as created by another particle, so it covers that case as well. A frequently used approximation to help ease of calculation is the Born approximation. It just so happens that even without the Born approximation, for a coloumb potential (which is any regular atom), the scattering problem is exactly solvable. And that solution yields the same results as Rutherford scattering. So quantum mechanically as well, everything is quite well explained.
EDIT 2: I'd earlier said that Rayleigh scattering was exactly solvable, but it was actually Rutherford scattering. Rayleigh scattering is a limit of Mie scattering which is valid for EM waves scattering of particles. However, I suspect that if you consider photons as particles that interact with the coulomb potential, rutherford scattering will be a valid quantum mechanical description of the phenomenon. I haven't been able to find any links with the relevant details worked out, but once I do I'll link them here.
More than one photon can be absorbed, but the probability is minute for usual intensities. As a scale for "usual intensities" note that sunlight on earth has an intensity of about $1000\,\mathrm{W/m^2} = 10^{-1}\mathrm{W/cm^2}$.
The intuitive reason is, that the linear process (an electron absorbs one photon) is more or less "unlikely" (as the coupling between the em. field and electrons is rather weak), so a process where two photons interact is "unlikely"$^2$ and thus strongly suppressed. So for small intensities the linear process will dominate distinctly. The question is only, at what intensities the second order effects will become visible.
In the paper by Richard L. Smith, "Two-Photon Photoelectric Effect", Phys. Rev. 128, 2225 (1962) the photocurrent for radiation above half of the cutoff frequency but below the cutoff frequency is discussed. They note that for usual intensities the photocurrent will be minute, but that given strong enough fields
such as those observed in a focus spot of a laser (on the order of $10^7\,\mathrm{W/cm^2}$) the effect might be measurable. They also note, that thermal heating by the laser field may make the pure second order effect unobservable.
The more recent paper S. Varró, E. Elotzky, "The multiphoton photo-effect and
harmonic generation at
metal surfaces", J. Phys. D: Appl. Phys. 30, 3071 (1997) dicusses the case where high intensities (on the scale of $10^{10}\,\mathrm{W/cm^2}$) produce even higher order effects (and unexpectedly high, coherent non-linear effects, that is absorption of more than two photons by one electron). Their calculations explain the experimental observations of sharp features in the emission spectra of metal surfaces.
Historical fun fact: The 1962 paper is so old, that it talks about an "optical ruby maser"; lasers where so new back then, they did not even have their name yet.
Best Answer
There are several different things that need to be explained / explored here.
First - the speed of light in vacuum is independent of frequency / wavelength. The same is not necessarily true for light in any medium other than vacuum: this is why we can see rainbows!
Second - not all objects emit "white" light. The emission spectrum of a star depends, among other things, on its temperature and composition. If the emission is mostly due to the temperature of the start, then you get black body radiation (yes, a black body can look white, red, yellow...). The black body radiation of an object depends on its temperature - the hotter it is, the more radiation there will be, and the more the spectrum shifts towards blue. If a star runs out of fuel but is not massive enough to become a supernova, it may become a red giant instead - big, "cold", red. The emission spectrum of a star may be further modified by the presence of certain atomic species - this can give rise to absorption or emission peaks in the otherwise smooth black body spectrum.
Third - if an object is moving towards the observer, the frequency of light observed will increase (blue shift); when it is moving away, you get red shift. Since the universe is expanding, objects that are far away exhibit greater red shift.
Fourth - as light travels through space, it will interact with (and be absorbed by) interstellar dust. The density may be very low, but when you have a long way to travel, it can add up. Actually, this is how NASA looks for potential stars-with-planets: if planets can form around a star, there is usually a lot of dust near that star; this dust will absorb some of the light from the star, and may re-emit it. But since the dust is much cooler than the star, this will change the spectrum towards the red (cooler).
Fifth - there is a particular mechanism for light scattering called Rayleigh scatter. The probability of light being scattered by a small particle is related to the size of the particle relative to the wavelength of the light - the shorter the wavelength, the stronger the scatter. This has two noticeable effects on earth: the sky is blue, and the sunset is red. That's really the same physics giving rise to different colors! When the sun is close to the horizon, the sunlight has to travel through a lot of the atmosphere to get to your eye. This means it will encounter a lot of small particles in the air, and that means that a lot of the blue light in the sunlight has a chance of being scattered. If you starts with all the colors and you scatter the blue, you are left with something that looks reddish. At the same time, the sky is blue because if you are not looking directly at the sun, then you are normally looking at "the black of space". Except that there is atmosphere in the way, and little particles in the atmosphere can give rise to scatter of sunlight (during the day). And since the most likely (visible) color to scatter is blue, this makes the sky blue.
Sixth - the perception of color. When you look at an object, it is really hard to know what color it is. Instead, you tend to judge the color based on what is nearby. When you are indoors, and the incandescent (tungsten) light is on, a piece of white paper will look "white" to you. If you use a different light source (say the sun light), the paper will still look white. But in fact it will be a very different color! Digital cameras can really show you this - they use something called "White Balance" to try to correct for the color of the light, but if you don't do that, an indoor photo will look very yellow. In the "olden days" of color film, you would use a special "tungsten filter" on your camera to correct for the yellow color; otherwise your photos would look horrible. And this explains your TV. If you are close to a TV, it looks bright, and white is white. As you walk further away, you see other things that are more yellow (because they are illuminated with incandescent light). By comparison, the "white" from the TV will look blue.
I wrote an earlier answer about the "color" of the moon; in that, I created a simple grayscale image that shows how our perception of one shade depends on what else we see. While it was not designed to demonstrate the "white is blue" thing, it might be instructive:
It is reasonable to say that the little rectangle on the left looks white, while it looks gray on the right; and whether you consider it gray or white in the middle depends on your screen brightness, probably. But they are in fact all the exact same shade of gray.