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?
It's difficult to explain this without using Lagrangians. So I'll just do my best to explain what the terms in a Lagrangian mean.
short answer: Left-handed particles feel whatever force fields they couple to in the Lagrangian.
For the sake of this discussion, we will define the Lagrangian as a mathematical object that completely governs the laws of motion of the particles therein. It will take many definitions, but I will do my best to be clear.
Definitions:
force fields: You will often see these referred to as gauge fields. They are the fields that describe the mediators of a force (e.g. the photon field $A_\mu$ which mediates the EM force).
feel: We usually say that a given particle feels a force if it couples to the corresponding force field.
coupling: For the sake of this discussion (but is actually quite general), we say that particles couple to a field if they are multiplied together in the Lagrangian.
Explanation
Now that the definitions are out of the way, lets get to the meat. The weak force is much harder to explain at a first pass, so what I will do is give the desired explanation first in terms of the electromagnetic force (often times called QED). The concepts will of course generalize.
QED
As I said before, the Lagrangian governs everything so the first thing we must write down is this. Note: writing down the correct Lagrangian for a given theory is not obvious and takes a tremendous amount of guesswork, so let's just say that we've done the guess work right.
We will represent an electron as $\Psi$ and an anti-electron as $\bar{\Psi}$, and we will represent the EM field by $A_\mu$. The QED Lagrangian is
$$ L = \bar{\Psi}(i\gamma^\mu \partial_\mu - m)\Psi - ieA_\mu \bar{\Psi} \gamma^\mu \Psi $$
Looks intimidating. However, based on our few rules we can already see that there is a coupling given by
$$ L_{cup} = - ieA_\mu \bar{\Psi} \gamma^\mu \Psi $$
What this means based on our above definitions is that electrons and anti electrons feel the electromagnetic force.
Handedness:
Now, it turns out all fermions (which we also represent by $\Psi$) can be written in the form
$$\Psi = \Psi_L + \Psi_R $$
I will not go into the how or why of this — we may just take it as a mathematical fact. So to find out if left-handed and right-handed fermions also feel the EM force, we plug this into our coupling Lagrangian and find that
$$ L_{cup} = - ieA_\mu \bar{\Psi}_R \gamma^\mu \Psi_R - ieA_\mu \bar{\Psi}_L \gamma^\mu \Psi_L - ieA_\mu \bar{\Psi}_L \gamma^\mu \Psi_R - ieA_\mu \bar{\Psi}_R \gamma^\mu \Psi_L$$
so we see that indeed both left-handed and right-handed $\Psi$ couple to the EM force (and hence feel it).
Relationship to Weak Force:
Now, I will not bother writing down the Lagrangian for the electroweak force as it will only cause further confusion. But I will mention this. The electroweak Lagrangian itself only contains couplings to fermions of the form
$$L \supset W_{\mu}\Psi_L \gamma^\mu \bar{\Psi}_L $$
where $W_\mu$ is the electroweak gauge field (force field). Again, this should not be obvious and coming up with the Lagrangian is very difficult. However, if we take this as fact (as we did before), we observe that only the $\Psi_L$s feel the weak force and the $\Psi_R$s, which do exist in the full Lagrangian, do not contain such a term. Hence they do not feel the weak force.
Best Answer
There are right handed quarks and leptons, they just don't interact weakly. The matter content of the Standard Model is often written in the "QUDLE" form as:
$$\begin{array} {} & SU(3)_C & SU(2)_L & U(1)_Y & U(1)_Q \\ \hline q_L^i = \begin{pmatrix} u_L^i \\ d_L^i \end{pmatrix} & 3 & 2 & 1/6 & \begin{pmatrix} \frac{2}{3} \\ -\frac{1}{3} \end{pmatrix} \\ \hline u_R^i & 3 & 1 & 2/3 & 2/3\\ \hline d_R^i & 3 & 1 & -1/3 & -1/3 \\ \hline l_L^i = \begin{pmatrix} \nu^i_L \\ e^i_L \end{pmatrix} & 1 & 2 & -1/2 & \begin{pmatrix} 0 \\ -1 \end{pmatrix} \\ \hline e_R^i & 1 & 1 & -1 & -1\\ \hline \phi & 1 & 2 & 1/2 & \begin{pmatrix} - \\ 0 \end{pmatrix} \\ \hline \end{array}$$
where it is understood that the Standard Model is a gauge theory with gauge group $SU(3) \times SU(2) \times U(1)_Y$ that gets spontaneously broken to $SU(3) \times U(1)_Q$. In particular, when people say that only left-handed quarks/leptons participate in weak force, they are talking about the fact that the up/down type left-handed quarks, as well as the left-handed leptons transform as the '2' representation under $SU(2)$, whereas the right-handed counterparts transform under the '1' (singlet) representation. This means that in the broken theory, the $W^\pm$ bosons which mediate decays don't interact with the right-handed quarks and leptons. (Note that the right-handed quarks and leptons still interact with Z bosons though, due to their $U(1)_Y$ charge)
The right handed quarks and leptons exist though. Note that after spontaneous symmetry breaking, the quarks and leptons gain mass terms that look like $m_u (\bar{u}_L u_R + \bar{u}_R u_L)$ (I am ignoring all the issues about flavor basis/mass basis not being same), these mass terms 'mix' the left and right handed modes, in the sense that you can't really treat them as separate.
In the table above, I also omit the neutrinos. Years ago, I believe people thought they could potentially be massless, and thus there could truly be no right-handed neutrinos. However, we have come to find out via experiment that neutrinos are massive, hence the simplest explanation would be that right-handed neutrinos do exist. I haven't included this in the table because