Special relativity was not born as a 4-covariant theory. Instead, Einstein derivated the kinematical quantities without spacetime,therefore without 4-vectors.
Consider then the following:
First Einstein derivated the kinematics of Relativity in terms of 3D Euclidean geometry and with the postulates of special relativity, showing the very fundamental concepts of physics in near to speed of light (e.g. time dilation). Then, Minkowski introduced the spacetime and invariant interval;the metric formalism of special relativity. Furthermore, with Minkowski's formalism a new type of object was needed to ensamble the kinematical quantities into spacetime formalism. One of these quantities was the velocity vector which required a concept of the 4-velocity vector. But in the transition to spacetime formalism, the velocity vector do not transforms (with a lorentz transformation) as a true vector . In order to construct the concept of velocity in spacetime we must consider proper time.
Well, this paragraph is my conclusion about the topic, but this is a intuitive conclusion. The formal explanation is then that 3-velocity is not a vector object, because the components didn't transforms as vectors under a lorentz transformation:
$$\frac{dx'^{\mu}}{dt'} = \Lambda^{\mu'}_{\nu}\frac{dx^{\nu}}{dt} \frac{dt}{dt'} = \frac{\Lambda^{\mu'}_{\nu}}{\Lambda^{0'}_{\nu}x^{\nu}}\frac{dx^{\nu}}{dt} \neq \Lambda^{\mu'}_{\nu}\frac{dx^{\nu}}{dt}\tag{1}$$
Even so, I still do not understand the following:
1) We can write the 4-velocity as, $$u(\tau) = (\frac{dx^{0}}{d\tau},\frac{dx^{1}}{d\tau},\frac{dx^{2}}{d\tau},\frac{dx^{3}}{d\tau})$$
but with relation,
$$d\tau = \frac{1}{\gamma}dt$$
we can write the 4-velocity in terms of 3-velocity $$u(t) = \gamma(c,\vec{v}(t))$$
I mean, if we already have a invariant object,the four velocity, why the one wants to consider 4-velocity in terms of 3-velocity?
