The two expressions aren't the same, I think. 1.5.3 assumes that ##\vec{\beta}_v=(\beta_v,0,0)^T##, while 1.5.4 makes no such assumption. You could just plug this assumption into 1.5.4 as a plausibility check. If you actually need to derive 1.5.4 then I'd start with four velocities and work from there, not from 1.5.3.
A slightly better plausibility argument is to argue that ##(\beta_v+\bar{\beta}_w^1,\bar\beta_w^2/\gamma_v,\bar\beta_w^3/\gamma_v)^T## could be said to be ##\vec\beta_v## plus the component of ##\vec{\bar\beta}_w## parallel to ##\vec\beta_v## plus ##1/\gamma_v## times the component of ##\vec{\bar\beta}_w## perpendicular to ##\vec\beta_v##.
The vector times the dot product in the last term in brackets in 1.5.4 pulls out the component of ##\vec{\bar\beta}_w## parallel to ##\vec\beta_v##, which is then added/subtracted appropriately to get what I wrote in words above.
The idea is to calculate the three-velocity ##\vec{w}## first for the simplifying case that ##\vec{v}=v \vec{e}_1##. Then one makes use of the fact that ##\vec{w}=\vec{W}/W^0## is a "three-vector", i.e., it transforms under rotations as a three-vector, and thus one can get the expression for an arbitrary ##\vec{v}## by writing (1.5.2) in a form that is kovariant under rotations; you can indeed check that when setting ##\vec{v}=v \vec{e}_1## in (1.5.3) you get back (1.5.2). Since (1.5.3) is written in a kovariant form under rotations, it must be correct for the general case, if it's correct for the special case.
