The bar notation is necessary in order to make all of the quantities above Lorentz invariant. In short the relationship between the bar'ed quantitites and charge conjugates of fermions have to do with taking the hermitian conjugate of the creation / annihilation opeartors.
Let us focus for now on a free theory of Fermions. A Dirac Fermion can be written down as the following solution to the Dirac equation
$$
\psi(x) = \sum_{s \pm}\int\tilde{dp}\big[ a_s(\textbf{p})u_s(p)e^{ip\cdot x}
+ b_s^{\dagger}(\textbf{p})\nu_s(p)e^{-ip\cdot x}\big] \\
\bar{\psi}(x) = \sum_{s \pm}\int\tilde{dp}\big[ a^{\dagger}_s(\textbf{p})\bar{u}_s(p)e^{ip\cdot x}
+ b_s(\textbf{p})\bar{\nu}_s(p)e^{-ip\cdot x}\big]
$$
where $\tilde{dp}$ is a Lorentz invariant measure and $u_s(p)$ and $\nu_s(p)$ are four-component spinors
$$
u_s(p) = \begin{pmatrix}
\sqrt{p\cdot \sigma} \;\xi^s \\ \sqrt{p\cdot \bar{\sigma}} \;\xi^s \end{pmatrix};\; \nu_s(p) =\begin{pmatrix}
\sqrt{p\cdot \sigma} \;\eta^s \\ -\sqrt{p\cdot \bar{\sigma}} \;\eta^s \end{pmatrix};\; \overset{(-)}{\sigma_{\mu}} = (\mathbb{1}, \pm \vec{\sigma})
$$
corresponding to fermion and anti-fermion spinors. $a^{(\dagger)}_s(\textbf{p})$ and $b^{(\dagger)}_s(\textbf{p})$ are annihilation (creation) operators that can act on the Vacuum
$$
a_s(\textbf{p})|0\rangle = b_s(\textbf{p})|0\rangle = 0 \\
a^{\dagger}_s(\textbf{p})|0\rangle = \frac{1}{\sqrt{2E_{\textbf{p}}}}|\textbf{p},s\rangle_{\text{ferm}};\;
b^{\dagger}_s(\textbf{p})|0\rangle = \frac{1}{\sqrt{2E_{\textbf{p}}}}|\textbf{p},s\rangle_{\text{anti-ferm}};
$$
Charge conjugation is a symmetry that is manifest in some theories of phenomenalogical interest (and some that are not). Therefore Charge conjugation is an operator that commutes with that Hamiltonion and thus, is a quantum number for states in the Hilbert Space.
For a general state one has
$\mathcal{C}|\phi\rangle = \eta_C|\phi^c\rangle$ for some field $\phi(x)$. Therefore, if we want to do a charge conjugation on some state, based off of how the creation operators act on the vacuum, what we want is
$$
\mathcal{C}^{-1}a_s(\textbf{p})\mathcal{C} = b_s(\textbf{p})
$$
and vice-versa. As it turns out, the charge conjugate of the Dirac fermion field operator is
$$
\mathcal{C}^{-1}\psi(x)\mathcal{C} = \psi^c(x) = \eta_cC\bar{\psi}^T(x).
$$
Noting the following from Srednickie pg 245
$$
C\bar{u}_s(p)^T = \nu_s(p);\; C\bar{\nu}_s(p)^T = u_s(p)
$$
one obtains
$$
\psi^c(x) = \sum_{\pm}\int\tilde{dp}\big[ b_s(\textbf{p})u_s(p)e^{ip\cdot x}
+ a_s^{\dagger}(\textbf{p})\nu_s(p)e^{-ip\cdot x}\big].
$$
Now both $\bar{\psi}$ and $\psi^c$ are different field operators. But think about how both operators act on the vacuum for this free Fermion theory.
To me at least this is how I see it. When people talk about anti-quarks or something in expressions that look like $\bar{u}(x)\gamma^{\mu}u(x)$, I just thing of it as an "abuse of notation" to make physical sense about the quantity we are looking at. In a way they are speaking about the "bar'ed" quantities with respect to how the corresponding free Fermion operator acts on the vacuum.
The reliability of this physical reasoning lies on the reliability of a perturbative expansion of some interacting theory over free fields. Strictly speaking, in strongly coupled theories (or even in something like low-energy QED if you are purist), this kind of physical argument has no rigorous basis.
I hope I answered your question. Hell I hope I am understanding this right. This is just the way I have always understood it.