Suppose we have Weyl spinor $\psi_{a}$, which transforms under irreducible representation $\left( \frac{1}{2}, 0\right)$ of the Lorentz group, $$ \psi_{a} \to (T(g))_{a}^{\ b}\psi_{b}, $$and complex conjugated spinor $\kappa^{\dot{b}}$, which transforms under $\left(0, \frac{1}{2}\right)$ representation, $$ \kappa^{\dot{b}} \to (T(g))^{\dot{b}}_{\ \dot{a}}\kappa^{\dot{a}} $$ Dirac spinor corresponds to the direct sum $\left( \frac{1}{2}, 0\right) \oplus \left( 0, \frac{1}{2}\right)$: $$ \Psi = \begin{pmatrix} \psi_{a} \\ \kappa^{\dot{b}}\end{pmatrix} $$ Charge conjugation of Dirac spinor is determined as $$ \tag 1 \Psi \to \hat{C}\Psi = \begin{pmatrix} \kappa_{a} \\ \psi^{\dot{b}}\end{pmatrix} $$ Sometimes charge conjugation is called "the Dirac version of of complex conjugation".
The question. Why do we need to introduce complex conjugation in a form $(1)$, not in a form of ordinary complex conjugation, $\Psi \to \Psi^{*}$?
This is a very good question. There are both mathematical and physical reasons for introducing the Dirac conjugate, which happen to converge on the same result.
Allow me to address your question head-on. First, I'll try to talk you out of considering standard conjugation $\Psi\mapsto \Psi^*$ where we conjugate each coefficient of $\Psi$ in some basis. This would be a perfectly respectable operation to do on a complex vector space, but the space of Dirac spinors is not $\textit{merely}$ a complex vector space. It also comes equipped with a representation. To draw an analogy, when we study smooth manifolds, generally smooth functions are all we care to talk about, even though it still makes sense to talk about continuous functions. Smooth functions just reflect the extra level of structure that has been added. When we work with representation spaces, we want the morphisms between vector spaces to complement the extra structure coming from the representations.
So there's a wispy philosophico-mathematical argument which I hope has begun to erode your confidence in complex conjugation in this context, but let's make things more precise. Suppose we have vector spaces $V, W$ carrying representations of a group $G$ and $T:V\rightarrow W$ is a linear map. We want to decide what it should mean for $T$ to be considered as a map of representation spaces. For any $v\in V$, we have the image $Tv\in W$, which is acted on by the group $G$. The group acts on both $V$ and $W$, so $Tv$ could equally claim to feel the influence of the action coming from $V$, i.e. $Tv\mapsto T(gv)$, or the action coming from $W$, i.e. $Tv\mapsto g(Tv)$. Each of these claims seem valid, so we take the equality of the two, $$T(gv)=g(Tv), \forall g\in G, \forall v\in V$$ as the definition of a map between representation spaces, and we call $T$ an intertwining operator. If $T$ is an isomorphism then $T$ is called an equivalence of representations. You may note that, $\textit{post hoc}$, this is mathematically a quite natural definition to consider.
Does the simple conjugation map on Dirac spinors have this property? Well, no: $$(S\Psi)^*=S^*{\Psi}^*\ne S{\Psi}^* $$ for all matrices $S$ in the image of the $(\displaystyle\frac{1}{2},0)\oplus (0,\displaystyle\frac{1}{2})$ representation of $SL(2,\mathbb{C})$. So from the point of view of representation spaces, complex conjugation is not a natural operation to consider. It's a bit too simple or naive; we want math to be as simple as possible, but not simpler (to paraphrase Einstein).
By the way, the 2-spinor $\epsilon$-tensor fits into this viewpoint. (Caveat: The conventions in this subject are in a state of anarchy from what I can tell, so my conventions here are almost certain to be at odds with yours and everyone else.) 2-spinors are introduced as objects with upper indices in a vector space $V$ that transform like $$\psi^a\mapsto A^a{ }_b \psi^b$$ where $A$ is a matrix in $SL(2,\mathbb{C})$. (I take my 2-spinors in the fundamental representation to have upper indices, which seems to be the first departure of my notation from yours.) Because matrices in $SL(2,\mathbb{C})$ have determinant $1$, one can note that any matrix of the form $$\epsilon_y= \begin{pmatrix} 0 & -y \\ y & 0 \end{pmatrix}$$ satisfies $\epsilon_y A=(A^{-1})^T\epsilon_y$. The natural representation of $SL(2,\mathbb{C})$ on $V^*$ defined by $\kappa\mapsto \kappa\circ A^{-1}$ is manifested in components by left multiplication by $(A^{-1})^T$ so the previous equation says $\epsilon_y:V\rightarrow V^*$ intertwines these two representations. We give elements of $V^*$ lower indices, so the group action is $$\kappa_a\mapsto (A^{-1})^b{ }_a\kappa_b=((A^{-1})^T)_a{ }^b\kappa_b,$$ and then $\epsilon_y$ has two lower indices; it takes an upper index and lowers it by $$\psi^a\mapsto \psi_b=\psi^a (\epsilon_y)_{ab}=(\epsilon_y^T)_{ba}\psi^a.$$ One then makes a choice between $y=\pm 1$ and calls this $\epsilon$. One should be a bit sensitive about this matrix $\epsilon$, which can be thought of as a bilinear form or a linear transformation. The way I do things is that the linear transformation is defined by contracting the bilinear form on the first index (which you can see in my definition of "lowering" above), but then the matrix of the bilinear form and the matrix of the transformation are related by transposition (as indicated after the second $=$ sign in the "lowering" definition above). I tend to take $y=1$ as my definition of the linear transformation, which means $y=-1$ corresponds to the matrix of the bilinear form.
Now, after seeing why naive conjugation doesn't work, how do we go about constructing an appropriate notion of complex conjugation? One way to do this is to just follow the indices, which after all are just keeping track of what kind of representation space the vectors are living in. If $$\Psi= \begin{pmatrix} \psi^a \\ \kappa_{\dot{b}} \end{pmatrix} \in V\oplus {\bar{V}}^*$$ is a Dirac spinor, we can just swap the $\psi$'s and $\kappa$'s while raising and lowering: $$C\Psi=\begin{pmatrix} \kappa^{\dot{a}} \\ \psi_b \end{pmatrix}=\begin{pmatrix} 0 & \epsilon^{\dot{a}\dot{b}} \\ \epsilon_{ab} & 0 \end{pmatrix} \begin{pmatrix} \psi^a \\ \kappa_{\dot{b}} \end{pmatrix}\in {\bar{V}}\oplus V^*.$$ Applying complex conjugation of vector spaces to this, the result is $$\hat{C}\Psi=\begin{pmatrix} \kappa^{a} \\ \psi_{\dot{b}} \end{pmatrix}\in V\oplus {\bar{V}}^*$$ which is guaranteed to be an intertwining map since the $\epsilon$-tensors were constructed so as to intertwine between representations.
Another approach which works is to think about representations of the full Clifford algebra. If one chooses $\gamma$-matrices, a transformation $C$ can be found which satisfies $$C\gamma^i=\overline{\gamma^i}C$$ giving an equivalence between the fundamental representation $S$ and the complex conjugate representation $\bar{S}$ of the Clifford algebra. This immediately implies $$C\gamma^i\gamma^jC^{-1}=\overline{\gamma^i}\,\overline{\gamma^j}$$ which can be exponentiated to $$C\exp(\gamma^i\gamma^j)C^{-1}=\overline{\exp(\gamma^i\gamma^j)}.$$ These exponentials cover all the elements $A\in SL(2,\mathbb{C})$ acting as matrices in the $(\displaystyle\frac{1}{2},0)\oplus (0,\displaystyle\frac{1}{2})$ representation by $S(A)=A\oplus (A^{-1})^T$ on $V\oplus {\bar{V}}^*$. Then we can write the exponentiated equation as $CS(A)=\overline{S(A)}C$, where $C:V\oplus {\bar{V}}^*\mapsto \bar{V}\oplus V^*$. Compare this to $C$ above to see that they have the same function. If you do some computations you find that these two ideas line up.
Physicists do all of this by thinking about Lorentz invariance, which means that the objects they consider have to possess the correct transformation properties when they change coordinates. When you really look at the equations they write down to express this, it manifests mathematically as the equivalence of different representations, so by thinking about the physics they've hit upon the same idea. I won't say more because this answer has gone on long enough and there aren't any degrees in physics hanging on my wall.