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.