The short answer to the (original) title question of "Does redshifts/blueshifts occur on Earth?" is yes, of course it does.
So why don't you notice, right?
The thing to note is that redshift from relative motion is proportional to the relative velocity divided by the speed of light.
The speed of light is 300,000 kilometers per second.
The fastest macroscopic1 objects that people accelerate go a few tens of kilometers per second.
So the ratio is of order $10^{-4}$ for the fastest human driven artifacts in existence. That is a small change.
The business with the Bezold–Brücke isn't a physics effect at all. It is---just as the quote says---a change in perception. That is, it is related to how the human visual apparatus and interpretation works. I'm cettainly not an expert in that field but it appears to just be a way of defining what is meant by "red" and "blue".
1 Of course, particle accelerators send small things to speed that approach that of light very closely, indeed. And under the right circumstances it is possible to observe very high Doppler shifts from some of these particles. But we don't generally bother because there are even more striking relativistic effect at play.
You can derive the relativistic Doppler shift from the Lorentz transformations. Let's start in the frame of the moving rocket, and let's take two events corresponding to nodes in the emitted wave (i.e. 1/$f$). Then in the rocket's frame the two events are (0, 0) and ($\tau$, 0), where $\tau$ is the period of the radiated wave. To see what the period of the radiation is in our frame we just have to use the Lorentz transformations to transform these two spacetime points into our frame.
For simplicity we'll take our rest frame and the frame of the rocket to coincide at $t = 0$. This is convenient because then the first event is just (0, 0) in both frames. Now the Lorentz transformations tell us:
$$ t' = \gamma \left( t - \frac{vx}{c^2} \right ) $$
$$ x' = \gamma \left( x - vt \right) $$
If we're tranforming from the rocket's frame to ours, and the rocket is moving at velocity $v$ wrt us, then we have to put the velocity in as $-v$, and we're transforming the point ($\tau$, 0). Putting these in the Lorentz transformations we find that the point ($\tau$, 0) in the rocket's frame transforms to the point ($\gamma \tau$, $\gamma v \tau$) in our frame.
The last step is to note that if we're sitting at the origin in our frame the light from the event at ($\gamma \tau$, $\gamma v \tau$) takes a time $\gamma v \tau/c$ to reach us. So the time we see the second event is $\gamma \tau + \gamma v \tau/c$ and this is equal to the period of the radiation, $\tau'$ in our frame:
$$ \tau' = \gamma \tau + \gamma v \tau/c $$
We just need to rearrange this to get the usual formula. Noting that $f'$ = 1/$\tau'$ and $f$ = 1/$\tau$ we take the reciprocal of both sides to get:
$$ f' = f \frac{1}{\gamma(1 + v/c)} $$
To simplify this note that:
$$\begin{align}
\frac{1}{\gamma} &= \sqrt{1 - \frac{v^2}{c^2}} \\
&= \sqrt{(1 - \frac{v}{c})(1 + \frac{v}{c})}
\end{align}$$
and substituting this back in our expression for $f'$ we get:
$$\begin{align}
f' &= f \frac{\sqrt{(1 - v/c)(1 + v/c)}}{1 + v/c} \\
&= f \frac{\sqrt{(1 - v/c)}}{\sqrt{1 + v/c}} \\
&= f \sqrt{\frac{c - v}{c + v}}
\end{align}$$
and presto it's proved!
Best Answer
You should not be seeing those words used interchangably. They are totally different concepts.
The path length is how far the light travelled to get from one place to another. If there's a light bulb over there 10 m away, and you're standing here, the path length is 10 m. If a light is shone into a piece of optical fiber and 10 km of fiber is spooled up on the bench and the light comes out the other end, the path length is 10 km, even though the source and destination are only a few meters apart. Neither of these depend on what the wavelength of the light is.
The wavelength is, like you say, the distance between minima or maxima of the electric field in the wave. For things that we call "light" this is usually somewhere in the range of 100 nm to 10 um. If you make a 632.8 nm HeNe laser, the output wavelength will be 632.8 nm, no matter how long a path you propagate the beam through. If you shine the beam into a piece of glass, the wavelength changes (because the frequency remains the same but the propagation velocity slows down) but once it exits the glass again, the original wavelength is restored.
It means either the atmosphere is thinner around the equator, or something about the setup means the beam will be directed in a more vertical direction when the experiment is done at the equator.
It has nothing to do with the wavelength, and the wavelength is not the same thing as the path length.
Edit
For the case of spectral irradiance, the result depends on both path length and wavelength. Path length because if light from the sun passes through a taller column of atmosphere, more of it will be absorbed before reaching the ground. Wavelength because different molecules in the atmosphere will absorb different wavelengths more or less.
So you should actually have a plot of spectral irradiance vs wavelength at the equator, and at each latitude up to the poles, with different path lengths through the atmosphere at each latitude.