The perfect head on collision is a special case where we don't need to worry about any relative angles. We can solve it using Physics 1 conservation of momentum and energy, all in the lab frame.
For equal mass particles
$$m u_1 = m v_1 + m v_2 \implies u_1 = v_1 + v_2,$$
and
$$\frac{1}{2} m {u_1}^2 = \frac{1}{2} m {v_1}^2 + \frac{1}{2} m {v_2}^2 \implies {u_1}^2 = {v_1}^2 + {v_2}^2 .$$
We can square the momentum equation and find
$${u_1}^2 = {v_1}^2 + {v_2}^2 + 2 v_1 v_2 = {v_1}^2 + {v_2}^2.$$
So
$$2 v_1 v_2 = 0.$$
This means one of the particles has zero velocity after the collision. Either particle 1 passed right through particle 2 with no effect (aphysical), or particle 1 stops and transfers all of its momentum and energy to particle 2.
In the center of mass frame this would be scattering at $180^\circ$. The fact that the first particle stops must break one of M&T's assumptions. In this case the scattering angle of particle 1 $\theta$ is ill defined. What is the direction of motion for a particle that is stopped?
In billiards you see a ball bounce backwards, because it was spinning backwards before the collision (while sliding along the table). It stops. Angular momentum is conserved so it keeps spinning. Friction catches the spin on the table and it starts to roll backwards. If the ball is rolling on the table it will continue to move forward after the collision for the same reason.
I don't have the reputation to comment, but setting $\theta=0$ is not the problem, as another answer suggests:
$$\tan\xi=\cot 0\rightarrow\infty \implies \xi=\frac{\pi}{2}$$
For energy conservation, the directions of the vectors are not important, as energy is a scalar quantity. For the kinetic energy you can simply plug in everything you have in the text into the equation you stated - as long as the collision is elastic. The directions only matter for the conservation of momenta, this is
$$m_1v_{1i}+m_2v_{1i}=m_1v_{1f}+m_2v_{2f}\,,$$
where you need to take care of the directions of the vectors, i.e., the direction of the momenta.
Sometimes it is useful to combine conservation of energy and conservation of momenta to solve for unknown quantities. See for example https://en.wikipedia.org/wiki/Elastic_collision
But be careful - kinetic energy is in general not conserved for inelastic collisions.
Best Answer
In general, there is no solution to the outgoing velocities of the particles. You need 6 components (three components of velocities for each particles) but you have only 4 equations (three components from the conservation of momentum, one from the conservation of energy). There are 2 equations missing.
To resolve this problem, you have to define what happens microscopically during the collision. For example, you can use a known potential for the interaction between the two particles and derive the trajectories.