Say you have a gas of atoms moving around in a box with some velocity. If the laser is placed at one side of the box, the photons' frequencies will be blue-shifted w.r.t. the laser if they're moving toward it, and red-shifted if they're moving away. So if you tune the laser to just above an atomic transition, the laser light will be transparent to the red-shifted atoms, and absorbed by the blue-shifted ones.
When an atom absorbs a photon, it is slowed down, since the photon and the atom make a head-on collision (sort of; the point being the momentum vectors have components that oppose each other along the laser-atom axis). At some later time it will reemit the photon in some random direction. If it emits the photon in the direction opposite its velocity (backwards), its momentum will be boosted to the same value it had before the absorption. However, any other direction of emission will give it a boost that is <= the momentum lost through the absorption.
So to summarize, absorption will only occur for atoms moving toward the laser. Once absorbed, they will be slowed down. With a very high probability, the subsequent emission be in some direction such that the net effect of absorption + emission is a negative change in momentum. Since the doppler shift is less with lower velocities, you have to keep detuning the laser as you cool.
Edit: @QuantumAI, I somehow didn't see your answer when I wrote mine.
Edit 2: @Rococo gave a very nice, concise explanation. Since you're still confused, let's take a look at a concrete example.
Say your atom initially has momentum $\mathbf{p}_{at,0} = p_i\mathbf{\hat x}$. And say, for the sake of simplicity that the photons have momentum $\mathbf{p}_{ph} = -\frac{p_i}{4}\mathbf{\hat x}$. Here the photon is incident from $-\mathbf{x}$, yielding a head-on collision with the atom traveling in $\mathbf{x}$ direction.
The initial magnitude of the momentum of the atom is clearly
$$\|\mathbf{p}_{at,0}\| = \sqrt{p_i^2} = p_i$$
The momentum after the absorption is
$$\mathbf{p}_{at,1} = \frac{3p_i}{4}\mathbf{\hat x}$$
with magnitude
$$\|\mathbf{p}_{at,1}\| = \frac{3p_i}{4}$$
Let's say now the emission occurs in some other direction, measured by an angle $\theta$ w.r.t. the horizontal. The momentum has the same magnitude as the photon, but the direction is different.
$$\mathbf{p}_{ph}' = \frac{p_i}{4}\left[\cos{\theta}~\mathbf{\hat x} + \sin{\theta}~\mathbf{\hat y}\right]$$
The final momentum of the atom is now
$$\mathbf{p}_{at,2} = \frac{p_i}{4}\left[ (3 + \cos{\theta}) ~\mathbf{\hat x} + \sin{\theta} ~\mathbf{\hat y}\right]$$
and the magnitude is
$$\|\mathbf{p}_{at,2}\| = \sqrt{\frac{p_i^2}{16}\left[ (9 + 6\cos{\theta} + \cos^2{\theta}) + \sin^2{\theta} \right]} = \sqrt{\frac{p_i^2}{16}(10 + 6\cos{\theta})}$$
Since $\cos{\theta}$ takes on values between $-1$ and $1$, you can see that the magnitude can take on values between $\frac{p_i}{2}$ and $p_i$.
The bottom line here is that for any angle of emission $\neq 0$, corresponding to an emission in the direction opposite the atom's velocity, the magnitude of the final momentum is less than that of the initial. Note that because this example looks at the case of a perfect head-on collision, it is idealized and therefore fairly unrealistic. Nevertheless, even for a different angle of incidence the concept is the same. As mentioned in the comments, if you average over a large number of events, the net effect is a reduction in momentum.
If this is still not clear, please ask a specific question about what is unclear to you.
Best Answer
Some laser rangefinding uses a retroreflector, which will bounce the laser light back in the direction it came regardless of orientation.
Otherwise, lasers operate at a very specific frequency, so the signal/noise ratio only needs to be strong enough to be detectable at that frequency.
If you shine a normal laser pointer on a wall, even if the wall is pretty far away, you can see the spot it makes. That means your eye can pick out the reflected laser light. The electronics can be made better than your eye, so it's not too hard to see reflected laser light..