Sorry for the slow reply. I slowed down a bit, reviewed your question and I believe found a satisfying answer to your question. Sorry it took me 4 replies before I actually read things carefully and got it right

Hope I didn't cause you too much headache
Malamala said:
For example if we have a photon traveling in +z direction doing the excitation and the one emitted by stimulated emission goes in -z (if the emitted one is in the +z, the net force would be zero and we would have no effective force), for the whole atom-laser system, we are just as in the beginning (atom in the ground state and 2 quanta coherent in the field), but it seems that now the atom gained a momentum, equal to twice the momentum of the photon, and hence a kinetic energy and I am not sure where this is coming from.
Your example requires that you have two counterpropagating laser optical modes for the atom to interact with, since you have photons going in both +z and -z directions. The simplest case is a standing wave in z: $$\vec{E}(\vec{x},t) = \vec{E_0} (\cos(kz + \omega t) + \cos(-kz + \omega t)$$
So where does the momentum go? The +k photon mode loses one photon worth of momentum (that's -1 photons times ##+\hbar k##), the -k photon mode gains one photon worth of momentum (that's +1 times ##-\hbar k##), and the atom gains 2 photons worth of momentum in the +z direction, as you said. Overall, that's ##(-1)(+\hbar k) + (+1) (-\hbar k) + 2\hbar k = -2\hbar k + 2\hbar k = 0##.
The key here is that the atom can interact with multiple optical modes while staying in the ground state at the end of the day. If you want to think of it as a Raman transition, then it'd be like a "lambda" (two low energy states, one high energy state) system where the atomic excited state is the go-between. One photon ends up switching sides from the ##+z## mode to the ##-z## mode.
Malamala said:
I am not sure how we get energy conservation
I couldn't find an answer to this in Cohen-Tannoudji. Here's what I can say:
The total Hamiltonian of the atom-laser system is given by $$H = H_L + \frac{p^2}{2m}+ H_a + V$$ where ##H_L## is the Hamiltonian of the photons before interaction with the atom, ##\frac{p^2}{2m}## is the Hamiltonian associated with the kinetic energy of the atom, ##H_a## is the atomic structure Hamiltonian before interaction, and ##V## is the interaction term. Per the previous example, one photon goes from the +z mode to the -z mode with the same wavenumber k, so there is no change in energy associated with ##H_L## since the number of photons and their frequency hasn't changed. Also, the atom gained a momentum of ##+2\hbar k##, so if we assume the atom started at rest then ##\frac{p^2}{2m}## increases by ##\frac{4k^2}{2m}##. The energy associated with ##H_a## doesn't change because the atom stays in the ground state the whole time. By process of elimination, that leaves a change in the interaction term ##V##.
When the atom gains momentum ##2\hbar k##, there is an associated Doppler shift such that the detuning changes as $$\delta \rightarrow \delta_{\pm} = \delta \pm kv$$ where ##\delta_+## refers to the detuning of the beam that propagates in +z, and vice versa for ##\delta_-##. In this case, ##v = p/m## and ##k = p/\hbar##, so $$\delta_{\pm} = \delta \pm \frac{p^2}{m \hbar}$$. Throw in a factor of ##\hbar## and you see an energy scale of ##\hbar \delta_{\pm} = \hbar \delta \pm \frac{p^2}{2m}##. You can use plug this into the expression for the E-field of the standing wave, evaluate ##\langle V \rangle = \langle -d \cdot E\rangle##, and that energy shift will compensate for the change in kinetic energy. I'd do the derivation here, but unfortunately I don't have time to finish that right now. If you're still interested, let me know in a reply and I'll get around to it when I can.