Frankly I find this so-called pedagogical article quite unintelligible and fail to see what the author wanted to say about these two operations. I also can't make any sense of the "derivation" of 7.3 based on the chiral projection being a "numerical matrix" and therefore commuting with charge conjugation operator.
Moreso the remark:
Elaborate statements are even made to the effect that charge
conjugation changes chirality. This makes no sense whatsoever, and can be best seen with Weyl fields for which chirality is the same as helicity. Helicity involves spin and momentum, none of which changes under charge conjugation. Thus helicity is unaffected by charge conjugation, and so must be chirality.
is misleading too. It's right that the left-handed Weyl spinor corresponds to the left helicity particle. But it also describes right helicity antiparticle (that the author notes himself earlier!) E.g. for a long time we didn't know that the neutrino has a mass and therefore we could in principle describe it by a single Weyl spinor. But in addition to the left helicity neutrino the same spinor describes right helicity anti-neutrino.
You can't redefine it away because helicity corresponds to a specific value of the angular momentum and therefore is observable.
The charge conjugation transforms the particle into the antiparticle with the same helicity. You can't get the left helicity antiparticle from purely left-handed Weyl spinor! So it's no surprise that the charge conjugation converts left-handed Weyl spinor into the right-handed and vice versa.
This flip is also why the weak interaction term $i\psi_L^\dagger\gamma^0\gamma_\mu W^\mu\psi_L$ is not invariant under $\mathcal{C}$ transform. The spatial reflection also changes the chirality and thus it happens to be invariant under the combined $\mathcal{CP}$.
For the Dirac fermion all works perfectly for a straightforward definition of the charge conjugation that you denote as $\hat{\psi}$. In the Fock space this leads to $\mathcal{C}a_s^\dagger(p)\mathcal{C}^{-1}=b_s^\dagger(p)$ where $a_s^\dagger$ and $b_s^\dagger$ are creation operators for particle and antiparticle respectively with a helicity $s$. As expected in the massless limit the $a_s^\dagger$ and $b_s^\dagger$ act on different chiral components separately while $a_s^\dagger$ and $b_{-s}^\dagger$ on the same one. So as I said the charge conjugation changes chirality.
It also works perfectly for a Majorana fermion understood as a Dirac spinor with a reality condition $\psi=\hat{\psi}$. The reality condition leads to the identification $a_s^\dagger(p)=b_s^\dagger(p)$ which is exactly that the Majorana particle is the same as an antiparticle with the same helicity. You can't talk about chirality anymore even in the massless case because you tie the right component to the left by definition.
In summary I have no idea what the author of this article tried to say. All works fine for straightforward definition that is used by everyone and it does change chirality.
UPDATE
Note that before we discussed only the chirality of fields but not the states.
The common way to derive helicity=chirality in the massless case is done simply for the Dirac equation. The helicity operator is given by $\frac{\vec{p}}{E}\cdot\vec{\sigma}$. The Weyl equation takes the form,
\begin{equation*}
\vec{p}\cdot\vec{\sigma}\psi_{L,R}=\pm E\psi_{L,R}
\end{equation*}
Dividing it on $E$ we get that the helicity eigenstates exactly coincide with chirality eigenstates. We may conclude that the helicity operator equals $\gamma_5$ i.e. chirality.
Any solution restricted to the left component will get left helicity. However the solutions associated with antiparticles have negative energies. They are reinterpreted in such a way that their physical energy, momentum and spin (and thus helicity) change the sign. In the QFT this is achieved by stating that $\psi$ operator contains the annihilation operator for the particle but the creation operator for the antiparticle. Because in the case of fermions the operators anticommute e.g. in the spin operator,
\begin{equation*}
\hat{S}_k=\frac{1}{2}\int d^4x :\psi^\dagger \begin{pmatrix}\sigma_k&0\\0&\sigma_k\end{pmatrix} \psi:
\end{equation*}
we get that the antiparticle contribution has the right (i.e. opposite to the naive) sign.
Considering this if we define the chirality operator as $\int d^4x :\psi^\dagger \gamma_5 \psi:$ it will coincide with helicity operator also in QFT. So we should assign right chirality to the antiparticle described by left-handed Weyl spinor and it will not change under $\mathcal{C}$
In summary. The chirality of the field operator is changing. The chirality of the state is not.
