Why ${\rm SL(2,C)}$ if everything can be derived with ${\rm SO(3,1)}$ and ${\rm SU(2)}$? By showing that the complexified Lie algebra of the proper Lorentz group ${\rm SO(3,1)}$ is equivalent to two the direct sum of two complexified ${\rm SU(2)}$ Lie algebras $$\mathfrak{so}(3,1)_\mathbb{C}=\mathfrak{su}(2)_\mathbb{C}\oplus \mathfrak{su}(2)_\mathbb{C}.$$ With this at our disposal, it is possible to find all the irreducible representations of that algebra. 
With this technology, I do not need to know or mention anything about the group ${\rm SL(2, C)}$, its Lie algebra or its representations. The knowledge of ${\rm SO(3,1)}$ and ${\rm SU(2)}$ seems to be sufficient, I think.


*

*So why learn about ${\rm SL(2, C)}$ in the first place? It seems to me to be unnecessary. Please tell me if I am missing something on a conceptual level. 

 A: To put it simply, to know that the Lie group $SL(2,\mathbb{C})$ of $2\times 2$ complex matrices with unit determinant is (the double cover of) the restricted Lorentz group $SO^+(1,3;\mathbb{R})$ gives a simple and direct way to understand how the restricted Lorentz group can act on a Weyl spinor $\psi\in\mathbb{C}^2$. In contrast, the group action (on a Weyl spinor) is somewhat mysterious/less intuitive from the point of $4\times 4$ spacetime Lorentz transformations $\Lambda\in SO^+(1,3;\mathbb{R})$. 
Similarly, the subgroup $SU(2)\subseteq SL(2,\mathbb{C})$ is (the double cover of) the 3D rotation group $SO(3)\subseteq SO(1,3;\mathbb{R})$. 
There is a corresponding double-copy version for the complexified proper Lorentz group $SO(1,3;\mathbb{C})$. For more details, see this & this related Phys.SE posts. 
A: As far as the Fierz identities are concerned you are right. Turns out that usually we also want to define complex conjugation. For instance we want to preserve CPT, so the field content should be closed under CPT action. 
If $a=1,2$ is the index of the first $SU(2)$ and $\dot a=1,2$ the second $SU(2)$, for the case $SO(1,3)$ the complex conjugation switch the representations:
$$
(\chi^{a})^{*}=\bar\chi^{\dot a}
$$
Which breaks $SU(2)_{\mathbb{C}}\times SU(2)_{\mathbb{C}}$ down to $SL(2,\mathbb{C})$.
Different signatures will impose different reality conditions to the Lorentz generators, reducing the $SU(2)_{\mathbb{C}}\times SU(2)_{\mathbb{C}}$ to a subgroup. 
For signature $SO(4)$ we get 
$$
(\chi^{a})*=\bar\chi_{a},\qquad (\chi^{\dot a})^{*}=\bar\chi_{\dot a}
$$
which leads to the $SU(2)\times SU(2)$ subgroup, without the complexification.
For signature $SO(2,2)$ we get
$$
(\chi^{a})*=\bar\chi^{a},\qquad (\chi^{\dot a})^{*}=\bar\chi^{\dot a}
$$
which leads to the $SL(2,\mathbb{R})\times SL(2,\mathbb{R})$.
So if you are interested in imposing Majorana conditions on spinors you see that $SO(4)$ this is not possible since $\varepsilon^{ab}\chi_{b}=\chi^{a}$ implies that $\chi^{a}=0$. The minimal number of components in that case is two complex, or four real. 
For the $SO(1,3)$ the Majorana condition fix the spinor $\chi^{\dot a}$ in terms of $\chi^{a}$, or vice-versa. The minimal number of components in that case is two complex or four real. 
For $SO(2,2)$ it is possible to impose a Majorana condition in a single chiral spinor, reducing the number of components to a total of two real components.
