Do you encounter more photons (per unit time) when moving forwards at a constant velocity? Let's say you have rain hitting you evenly on all sides (not very realistic, I know). If you were to move forwards at a constant speed, there would be more droplets of rain hitting you per second on your front, since the relative speed of droplets in front of you has increased.
Now, if you were to have photons 'hitting' you evenly on all sides and you move forwards at a constant speed, surely the relative speed of the photons in front of you WON'T increase (since light travels at the same speed to all observers), and therefore photons will still be hitting you evenly on all sides (per second).
However the searchlight effect seems to disagree with my conclusion. What have I done wrong?
 A: You are right in that the speed of light doesn't change. It is a completely different effect to the rain drop analogy. If you had only light hitting you directly from the front and directly form the back, you would observe the same intensity in the moving frame (only blue/red shifted). But for light coming at you from an angle $\theta_s$ in the rest frame, the angle changes when you move with velocity $v$, the new angle in the moving frame is:
$ \theta_o=arccos(\frac{\cos \theta_s-\frac{v}{c}}{1-\frac{v}{c} \cos \theta_s})$
such that the angle shifts toward the front direction. And if you now had a uniform intensity in the rest frame, in the moving frame more light comes from the (general) forward direction and less from the (general) behind direction. Hope this helps. 
A: What happens if you first drive on a road at speed 0.99 c, and then accelerate to speed 0.999 c?
You will pass milestones at much faster rate. That happens mostly because of length contraction, not because of larger speed, as the speed increase was only 0.009 c.
Roughly the same contraction as above will happen to a line of photons approaching you, if you change your speed by 0.009 c towards the photons.
The exact amount of contraction is the same as the contraction of the wave-length of the photons, which can be calculated using the relativistic Doppler-shift formula.
A: Let's say Bob is standing still while a one light second long photon formation flies past him. How long does the passing of the photon formation and Bob take according to us? 
Answer: It takes one second. 1 light seconds / c = 1 seconds.
Let's say Jim is moving forwards at speed 0.1 while a one light second long photon formation moving to the opposite direction flies past him. How long does the passing of photon formation and Jim take according to us?
Answer: It takes 0.909 seconds. 1 light second / 1.1 c = 0.909 seconds. (Using relativistic velocity addition would be an error)
Because of time dilation according to Bob the passing takes a little bit less than 0.901 seconds. 
This is one way to calculate at what rate photons pass a moving observer, or collide with him.
