Once more, the confusion seems to be due to the use of old-fashioned ideas which were long abandoned in the physics community but still being used in popular-science writings and (which is a crime) even in some textbooks. One such things are the various relativistic masses. We had a long thread about it recently, which unfortunately ended up in the insistence of some posters to introduce even more nonsensical notions like norms which can get imaginary. I'll stick to the modern covariant formulation.
Masses are real scalar quantities. To emphasize that this one and only useful definition of mass is used, one sometimes call it invariant mass.
Another quantity which usually makes trouble is the "three velocity". This is somewhat harder to eliminate, because sometimes it's convenient to have it in the context of various density, where it can be used to build covariant densities from it, e.g., in any frame ##(c \rho(t,\vec{x}), \vec{v}(t,\vec{x}) \rho(t,\vec{x}))##, where ##\rho## is a charge density (as measured in the arbitrary frame) and ##\vec{v}## is the three-velocity-field of the charges (described in a continuum-mechanical manner, which is the natural description in any field theory). is a four-vector field. We come back to the fact that this is really a four-vector field in a moment.
Now consider first point charges in special relativity. You can describe its motion covariantly by a world line, ##x^{\mu}(\lambda)##, where ##\lambda## is a scalar parameter. Now the three-velocity is defined by ##\vec{v}=\mathrm{d} \vec{x}/\mathrm{d} t##, which makes it a complicated non-covariant object, because it uses the arbitrary components with respect to the chosen Minkowski basis, to which the components ##x^{\mu}## refer.
For massive particles it's much more convenient to choose the proper time of the particle, which is invariant as the world-line parameter. It's defined by
$$c^2 \mathrm{d} \tau^2=\eta_{\mu \nu} \mathrm{d} x^{\mu} \mathrm{d} x^{\nu}$$
and a scalar. Then the socalled four-velocity
$$U^{\mu}=c u^{\mu} = \mathrm{d}_{\tau} x^{\mu}$$
are four-vector components, which transform under Lorentz transformations as any four-vector. It's easy to see that the relation between the spatial part of this "proper velocity" and the "three-velocity" is given by
$$\vec{v}=\mathrm{d}_t \vec{x} = \frac{\mathrm{d} \tau}{\mathrm{d} t} \mathrm{d}_{\tau} \vec{x}=\sqrt{1-\vec{v}^2/c^2} \vec{U} \; \Rightarrow \; \vec{U}=\gamma \vec{v}, \quad \gamma=1/\sqrt{1-\vec{v}^2/c^2}.$$
You can also write
$$(U^{\mu})=\gamma (c,\vec{v}).$$
With this the velocity-addition law becomes easy to derive. Just first consider the proper velocity. Say in some reference frame ##\Sigma'## the proper velocity of a particle is ##U^{\prime '}=(c,v',0,0,0) \gamma_{v'}## and now let this frame move with a three-velocity ##\vec{w}## in ##x^1##-direction wrt. a reference frame ##\Sigma##. Since ##U^{\mu}## is a four-vector it transforms as any four-vector under the appropriate (rotation-free) Lorentz transformation,
$$U^0=\gamma_w (U^{\prime 0}+w/c U^{\prime 2}), \quad U^1 = \gamma_w (U^{\prime 0}+w/c U^{\prime 0}, \quad U^2=U^{\prime 2}, \quad U^{3}=U^{\prime 3}.$$
Now the three-velocity in ##\Sigma## is given by
$$\vec{v} = c \frac{\vec{U}}{U^0}.$$
Using the above Lorentz transformation result this gives
$$v^1=c \frac{U^1}{U^0}=\frac{ \gamma_w(c U^{\prime 1}+ w U^{\prime 0})}{\gamma_w (U^{\prime 0}+w/c U^{\prime 1})}.$$
Now dividing both numerator and denominator by ##U^{\prime 0}## yields
$$v^1=\frac{v'+w}{1+ v w/c^2}.$$
Of course in this case ##v^2=v^3=0##.
Now let's briefly come back to the above claim that ##(j^{\mu})=(c \rho,c \vec{v} \rho)## is a four vector. Writing it in terms of the four-velocity vield ##(U^{\mu}) = \gamma (c,\vec{v})## leads to
$$j^{\mu}=\frac{\rho}{\gamma} U^{\mu}=\rho_0 U^{\mu}.$$
it's clear that ##\rho_0=\rho/\gamma## is the charge density as measured in the instantaneous local rest frame of the charge-fluid cell. The charge is a scalar, but the volume of the fluid cell is Lorentz contracted in the frame, where this cell is moving, i.e.,
$$\mathrm{d} V=\mathrm{d} V_0/\gamma \; \Rightarrow \; \rho=\mathrm{d} Q/\mathrm{d} V = \mathrm{d} Q_0/V=\mathrm{d} Q_0/\mathrm{d} V^0 \gamma=\rho_0 \gamma.$$
One can also see this directly, because the four-current in the local rest frame is
$$(j^{\prime \mu})=(c \rho_0,0,0,0) \; \Rightarrow \; (j^{\mu})=\gamma \rho_0 (c,\vec{v})=\rho(c,\vec{v}).$$
