The charged current part of the Lagrangian of the electoweak interaction, for the first generation of leptons, is :
$$L_c = \frac{g}{\sqrt{2}}(\bar \nu_L \gamma^\mu e_L W^+_\mu + \bar e_L \gamma^\mu \nu_L W^-_\mu )$$
The first part corresponds to different versions of the same vertex :
$e_L + W^+ \leftrightarrow \nu_L \tag{1a}$
$(\bar\nu)_R + W^+ \leftrightarrow(\bar e)_R \tag{1b}$
$W^+ \leftrightarrow (\bar e)_R +\nu_L \tag{1c}$
The second part corresponds to different versions of the hermitian congugate vertex :
$\nu_L + W^- \leftrightarrow e_L \tag{2a}$
$ (\bar e)_R + W^- \leftrightarrow(\bar \nu)_R \tag{2b}$
$W^- \leftrightarrow e_L +(\bar \nu)_R \tag{2c}$
Here, $(\bar e)_R$ and $(\bar\nu)_R$ are the anti-particle of $e_L$ and $\nu_L$
Roughly speaking, you can change the side of a particle relatively to the $\leftrightarrow$, if you take the anti-particle.
Why the right-handed particles appear ? The fundamental reason is that we cannot separate particles and anti-particles, for instance, we cannot separate the creation of a particle and the destruction of an anti-particle.
[EDIT]
(Precisions due to OP comments)
The quantized Dirac field may be written :
$$\psi(x) = \int \frac{d^3p}{(2\pi)^\frac{3}{2} (\frac{E_p}{m})^\frac{1}{2}}~\sum_s(b(p,s) u(p,s)e^{-ip.x} + d^+(p,s) v(p,s)e^{+ip.x} )$$
$$\psi^*(x) = \int \frac{d^3p}{(2\pi)^\frac{3}{2} (\frac{E_p}{m})^\frac{1}{2}}~\sum_s(b^+(p,s) \bar u(p,s)e^{+ip.x} + d(p,s) \bar v(p,s)e^{-ip.x} )$$
Here, the $u$ and $v$ are spinors corresponding to particle and anti-particle, the $b$ and $b^+$ are particle creation and anihilation operators, the $d$ and $d^+$ are anti-particle creation and anihilation operators.
We see, that in Fourier modes of the Dirac quantized field, the elementary freedom degree is (below $p$ and $s$ are fixed):
$$b(p,s) u(p,s)e^{-ip.x} + d^+(p,s) v(p,s)e^{+ip.x}$$
Now, suppose we are considering massless particles, so that helicity and chirality are the same thing. Suppose that, for the particle (spinor $u(p,s)$) the couple $s,p$ corresponds to some helicity. We see, that, for the anti-particle ($v$), there is a term $e^{+ip.x}$ instead of $e^{-ip.x}$ for the particle. That means that the considered momentum is $-p$ for the anti-particle, while the considered momentum is $p$ for the particle. The momenta are opposed for a same $s$, so it means that the helicities are opposed.
You are correct that for a massive spinor, helicity is not Lorentz invariant. For a massless spinor, helicity is Lorentz invariant and coincides with chirality. Chirality is always Lorentz invariant.
Helicity defined
$$
\hat h = \vec\Sigma \cdot \hat p,
$$
commutes with the Hamiltonian,
$$
[\hat h, H] = 0,
$$
but is clearly not Lorentz invariant, because it contains a dot product of a three-momentum.
Chirality defined
$$
\gamma_5 = i\gamma_0 \ldots \gamma_3,
$$
is Lorentz invariant, but does not commute with the Hamiltonian,
$$
[\gamma_5, H] \propto m
$$
because a mass term mixes chirality, $m\bar\psi_L\psi_R$. If $m=0$, you can show from the massless Dirac equation that $\gamma_5 = \hat h$ when acting on a spinor.
Your second answer is closest to the truth:
The weak interaction couples only with left chiral spinors and is not frame/observer dependent.
A left chiral spinor can be written
$$
\psi_L = \frac12 (1+\gamma_5) \psi.
$$
If $m=0$, the left and right chiral parts of a spinor are independent. They obey separate Dirac equations.
If $m\neq0$, the mass states $\psi$,
$$
m(\bar\psi_R \psi_L + \bar\psi_L \psi_R) = m\bar\psi\psi\\
\psi = \psi_L + \psi_R
$$
are not equal to the interaction states, $\psi_L$ and $\psi_R$. There is a single Dirac equation for $\psi$ that is not separable into two equations of motion (one for $\psi_R$ and one for $\psi_L$).
If an electron, say, is propagating freely, it is a mass eigenstate, with both left and right chiral parts propagating.
Best Answer
The article you link is choosing to use non-standard terminology to explain how the weak force can couple to particles with a certain handedness but not others. I do something similar with "electron-1" and "electron-2" particles in this answer of mine - it's a pedagogic trick, not a claim about things being "really" different particles.
Standard usage is to call both the left-handed and the right-handed versions of a negatively charged lepton with the mass of an electron "electron", and both the left-handed and the right-handed versions of a positively charged lepton with the mass of an electron "positron". Since the electron is massive, the two versions with different handedness couple to each other - the equations of motion for a massive Dirac field mean that a left-handed solution does not stay purely left-handed and likewise for the right-handed versions. So, ordinarily, it is natural to not view the states of different handedness as "different particles" since they almost always occur in a mixture anyway. Therefore, standard usage only speaks of "the electron" and "the positron" - a particle that evolves into another particle just by being left alone (with no decay products or anything) isn't usually what we think of as a distinct particle.
In the context of the weak force (or any other hypothetical chiral interaction) however, it is only the parts with a certain handedness that interact, and since we usually say that particles either interact or don't interact (and not that "one half of a particle interacts" or whatever), in this context it can be useful to conceptualize the chiral parts of the full Dirac electron as "different particles". This doesn't change anything about what real electrons and positrons are (namely almost always mixtures of these two chiral states), it's just a different way of speaking about them.