In your second paragraph, you are viewing $SL(2,\mathbb C)$ as a real Lie group. Its Lie algebra is $\mathfrak{sl}_2(\mathbb C)$ viewed as a real Lie algebra. The complexification is therefore $\mathfrak{sl}_2(\mathbb C)\otimes_{\mathbb R}\mathbb C$. This is not the complex Lie algebra $\mathfrak{sl}_2(\mathbb C)$: for one thing, its dimension is $6$ while the latter has dimension $3$.
Now, if $g$ is a complex Lie algebra and we let $g_{\mathbb R}$ be $g$ viewed as a real Lie algebra, wwe always have that the complexification of $g_{\mathbb R}$ is $g\oplus g$. Proving this is an instructive exercise.
I have been thinking about this the past few days in preparation for an exam at EPFL as a result of some really shitty course notes. My familiarity with the subject is thus rather poor but at least I sympathize with your plight for clarity.
1 . I think that key to working with this problem is to first make concrete what the complexification of $\mathfrak{su}(2)$, $\mathfrak{su}(2)_\mathbb{C}$, really is and what its algebra is. We know that the natural basis of the $\mathfrak{su}(2)$ are the Pauli matrices $\{\sigma_1, \sigma_2, \sigma_3\}$ with the familiar Lie Bracket $[\sigma_i, \sigma_j] = i \varepsilon_{ijk}\sigma_k$. This is a REAL vector space and the complexification is a particular complex vector space where the Lie bracket is essentially what we expect it to be when treating the bracket as if it is linear over $i$ as well
$\mathfrak{su}(2)_\mathbb{C}$ is the Lie algebra of formal sums $u + iv$ where $u,v \in \mathfrak{su}(2)$ and where the complexified Lie-bracket expressed in terms of the real Lie bracket is
$$[x + iy, u + iv]_{\mathbb{C}} = ([x,u] - [y,v]) + i([x,v] + [y,u])$$
I wont write the complex sign again as its easy to take as implicit. Now that we hopefully agree on the definition I am probably going to annoy you by viewing complexified algebras as real algebras of twice the dimension because I find this situation to be more transparent. I am free t view my complexified algbra as a real algebra and in this picture the most natural basis we can come up with is
$$\sigma_1, \sigma_2, \sigma_3, i \sigma_1, i\sigma_2, i\sigma_3$$
I check the resulting Lie brackets and we end up with
$$[\sigma_i, \sigma_j] = i \varepsilon_{ijk}\sigma_k \\ [\sigma_i, i\sigma_j] = i \varepsilon_{ijk}(i\sigma_k) \\ [i\sigma_i, i \sigma_j ] = -i \varepsilon_{ijk}\sigma_k$$
We easily see a correspondence
$$J_j \leftrightarrow \sigma_j \qquad K_j\leftrightarrow i\sigma_j$$
and conclude
$$\mathfrak{so}(1,3) \simeq \mathfrak{su}(2)_\mathbb{C}$$
thus to be it looks like it is the REAL $\mathfrak{so}(1,3)$ which is isomorphic to the complexification of $\mathfrak{su}(2)$ (but also viewed as a REAL Lie algbera, of real dimension $6$). I find this to be a much more transparent way of arriving at the isomorphism rather than going via the complexification.
2. To me this looks like it will imply
$$\mathfrak{so}(1,3)_\mathbb{C} \simeq (\mathfrak{su}(2)_\mathbb{C})_\mathbb{C} \simeq \mathfrak{su}(2)_\mathbb{C} \oplus_\mathbb{C}\mathfrak{su}(2)_\mathbb{C} $$
I have to admit I don't know how to make sense of going via the complexification of $\mathfrak{so}(1,3)$ neither. I had an argument planned out but it collapsed and I reverted to the one above. Maby I'll try to fix this if you come back and discuss it with me.
3. I started thinking about this but I think you actually mean $\mathfrak{sl}(2,\mathbb{R})_\mathbb{C} \simeq \mathfrak{sl}(2,\mathbb{C}) \simeq \mathfrak{su}(2)_\mathbb{C}$? $\mathfrak{sl}(2,\mathbb{C})$ is a real vector space made up of traceless complex matrices so the 6 most obvious basis matrices are
$$\alpha_1 = \begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}, \alpha_2 = \begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}, \alpha_3 = \begin{pmatrix}0& 0 \\ 1 & 0\end{pmatrix}, \; \text{and} \; i\alpha_1, i\alpha_2, i\alpha_3$$
From this we can find an explicit change of basis to the complexified Pauli matrices $$\sigma_1 = \alpha_2 + \alpha_3, \quad \sigma_2 = i\alpha_1 - i\alpha_3, \quad \sigma_3 = \alpha_1\\
i \sigma_1 = i\alpha_2 + i\alpha_3, \quad i\sigma_2 = \alpha_1 - \alpha_3, \quad i\sigma_3 = i\alpha_1$$
and since the the bracket is the commutator we see that the Lie-structures of these two Lie algebras are the same meaning they are the same.
4. To me it looks like we will have $\mathfrak{so}(1,3) \simeq \mathfrak{sl}(2,\mathbb{C})$ (where the latter is viewed as a $6$-dimensional real Lie algbera) which kind of surprises me.
5. Well if 4. holds then it should hold.
Best Answer
Maybe this is more of a comment but a general method to show that a Lie group $G$ is isomorphic to $SO^+(p,q)$ is to find a $p+q$ dimensional representation of $G$ that preserves an inner product of signature $(p,q)$. Then if $\dim G = \dim SO(p,q)$ and $G$ is connected, this will give an isomorphism $G/\ker \to SO^+(p,q)$.
In the case of $G = SU(2)$ and $SO(3)$, we need a 3-dimensional rep of $SU(2)$. Since $SU(2)$ is three dimensional, we can try the adjoint representation of $SU(2)$ on its Lie algebra. Since $SU(2)$ is compact any real representation is orthogonal so this maps into $SO(3)$. Then we just need to check that the kernel is $\{\pm 1\}$. Note also here that there is no need to go to the Lie algebra.
For your case of $SL(2,\mathbb C)$ the adjoint representation is 6 dimensional and irreducible since $SL(2,\mathbb C)$ is simple.