Why does reduced mass help when talking about two body problems?
You're transforming a problem in terms of two particles with individual masses and accelerations into a problem with a single fictitious particle. That simplifies the problem and lets you separate the relative motions and interactions of the particles from overall translation of their center of mass. 

In step 3, you need the relative acceleration $a_{rel}$ because that's the acceleration the single fictitious particle has.

In step 4, you're calculating the mass $\mu$ of the fictitious particle.

Before, you had $F_{12} = m_1 a_1$ and $F_{21} = m_2 a_2$ (two forces, two masses, two accelerations). Now you've got $F_{12} = \mu \ a_{rel}$ (one force, one mass, one acceleration). The price is that we've thrown away information about the overall translation of the system through space; we're only looking at the relative motions of the two particles. 

The coordinates of the fictitious particle are relative or internal coordinates $$x = x_2-x_1,\quad y = y_2-y_1, \quad z = z_2-z_1$$
This is useful because the potential energy $V$ of the interacting particles depends only on these relative coordinates. Making use of reduced masses and relative velocities and coordinates cleanly separates internal motion from translation of the system as a whole. The classical energy of the two body system will be
$$E = \left[\frac{1}{2\mu}(p_x^2+p_y^2+p_z^2)+V(x,y,z)\right] + \left[\frac{1}{2M}(p_X^2+p_Y^2+p_Z^2)\right]$$
where $p_i$ means momentum for motion along the $i$-th coordinate. The first term in brackets is the energy of our fictitious particle with mass $\mu$. The second term is the energy of a second fictitious particle with total mass $M=m_1+m_2$ and coordinates  $X$, $Y$, and $Z$ located at the center of mass of the system (e. g. $X = (m_1x_1 + m_2x_2)/M$). 

In chemistry, we don't care about the second term. For example, in an atom, we'll assume that the nucleus is clamped in place so we can just look at interactions within the atom.