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
Rant and introduction
There are a lot of wrong statements in lecture notes about chirality and helicity, and plenty of wrong answers on this site as well. I've personally written too many answers on this subject (1, 2, 3, 4), but here's a fifth that lays out the ideas again, in case people want it for future reference.
Therefore, "the chirality of a particle" is meaningless. If you have a left-helicity massless particle, for example, it could be annihilated by a left-chiral Weyl spinor field or created by a right-chiral Weyl spinor field. The particle doesn't care which kinds of fields you use to describe it. Any Lagrangian containing Weyl spinors, for instance, can be trivially rewritten in terms of only left-chiral or right-chiral Weyl spinors (or any mix of the two) by sprinkling in complex conjugates.
The statement "chirality is the same thing as helicity for massless particles" is also meaningless. When people say this, they're trying to convey that a massless left/right-chiral field annihilates a left/right-helicity particle. But it also creates a particle of the opposite helicity. That's the origin of your second confusing statement, "left-chiral anti-particles are right-handed", which contains an honestly impressive number of mistakes in just seven words. The closest correct analogue of this statement is "a right-chiral field creates left-helicity particles with the opposite internal quantum numbers".
Why do people seem to get this wrong, time and time again? One reason is that these arguments are not "load bearing". Many people know how to calculate cross sections, and you don't really need to understand the difference between chirality and helicity to turn the crank correctly. At the end you get the expected result, so there's no need to think harder. Another reason is that often, people who understand the difference perfectly well will elide the distinction between particles and fields to make the subject easier for new students. (I don't think it helps though, because it tends to get them very confused later, losing time on net.)
Answering the question
Anyway, with that rant out of the way, let's focus on your example. For simplicity, let's suppose the electron is massless. Let's focus on the part of the current you pointed out, $$\bar{e}_L \gamma^\mu e_L.$$ In this notation, $e$ is a four-component Dirac spinor field, $e_L$ is the left-chiral part, and $\bar{e}_L$ is its conjugate. Therefore, using the rules laid out above,
We want to use this vertex to create two particles, so it must create a right-helicity positron and a left-helicity electron. But this violates angular momentum conservation. You can also run the same argument with the other part of the current $\bar{e}_R \gamma^\mu e_R$, in which case the helicities flip and the same conclusion applies. Therefore, the process cannot occur for a massless electron, so it is helicity suppressed. (In reality, the decay $\pi^0 \to e^+ e^-$ occurs through a loop diagram with two copies of the electromagnetic vertex, but the same logic applies.)