I think you are sort of reversing the logic of chirality and helicity in the massless limit. Chirality defines which representation of the lorentz group your Weyl spinors transform in. It doesn't 'become' helicity, helicity 'becomes' chirality in the massless limit. That is, chirality is what it is, and it defines a representation of a group and that can't change. This other thing we have defined called helicity just happens to be the same thing in a particular limit.
Now once you take the massless limit the Weyl fermions are traveling at the speed of light you can no longer boost to a frame that switches the helicity. I think its best to think of a fermion mass term as an interaction in this case and remember that the massive term of a Dirac fermion is a bunch of left and right- handed Weyl guys bumping up into one another along the way. Conversely if you want to talk about a full massive Dirac fermion that travels less than c and you can boost to change the helicity, but that full Dirac fermion isn't the thing carrying weak charge, only a 'piece' of it is.
See this blog post on helicity and chirality.
As far as the left-right symmetry being broken people have certainly built models along these lines but I don't think they have worked out.
Does this answer your question?
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.
Best Answer
I would strongly disfavor your associating handedness with helicity instead of chirality, but no matter... Indeed, the WP usage of right-handed for right-chiral (annihilated by $1-\gamma_5$) is fine. It is a source of clarity, not confusion. In this issue, helicity stays out of the picture, regardless of language.
Right-chiral fermion fields have a hypercharge proportional to their electric charge, by the (EWeak) Gell-Mann—Nishijima formula. They do interact with the Higgs field, in the Yukawa terms of vanishing total weak hypercharge. These terms couple right-chiral to left-chiral fermions and induce the fermion masses through the v.e.v. of the Higgs field, of course. (Charged current weak interactions refer to W couplings, and not Higgs couplings.)
Right-chiral electrons exist, and they are, on average, half the degrees of freedom my and your and the universe's electrons have.
They are never isolated, as the electron mass connects them to their left-chiral brothers. Chirality is not constant in time. They have little to do with the Higgs mechanism. (I hope you don't actually mean SSB, instead).
NB There is a pedagogical picture presenting this as a chirality oscillation controlled by the mass term in the hamiltonian. A stationary solution of the free Dirac equation dictates immediately, starting from the R state, $$ \langle \gamma_5 \rangle = \cos(2mt). $$