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.
We define an inner product over $\Bbb C^{n \times n}$ by $\frac 1n \langle A,B \rangle = \operatorname{tr}(A^\dagger B)$; this is (a normalized version of what is) known as the "Frobenius" or "Hilbert-Schmidt" inner-product.
Note that for any $U$, the matrices $U\sigma_j U^\dagger$ form an orthonormal basis for the space of trace free $2 \times 2$ matrices (if you like, the orthogonal complement of the span of $I$). That is, we have
$$
\langle U\sigma_jU^\dagger,U\sigma_kU^\dagger \rangle = \delta_{jk}
$$
where $\delta_{jk}$ is a Kronecker-delta, and every trace-zero matrix can be written as a linear combination of these matrices.
The matrix $B_{ij}$ that you describe is the change-of-basis matrix that takes us from a coordinate-vector relative to the basis $\{U\sigma_jU^\dagger: j =1,2,3\}$ to a coordinate vector relative to the basis $\{\sigma_j: j = 1,2,3\}$. Because where are changing between two orthonormal bases, the resulting change-of-basis matrix is unitary.
Best Answer
Looking at a general basis will not help much, I guess, but what you have done can be completed: Note that the subspace $U = \def\<#1>{\left<#1\right>}\<A_1, A_2>$ of $\def\sl{\mathfrak{sl}(2, \mathbf R)}\sl$ generated by $A_1$ and $A_2$ is a subalgebra, as $[A_1, A_2] = A_2 \in U$. So $\sl$ has a two-dimensional subalgebra. We will prove that $\def\su{\mathfrak{su}(2,\mathbf C)}\su$ does not have any. Define $\phi \colon \su \to \mathbf R^3$ by $\phi(\sigma_i) = e_i$, where $e_i$ denotes the $i$-th standard unit vector. Then we have $$ \phi([\sigma_i, \sigma_j]) = \phi(\epsilon_{ijk}\sigma_k)=\epsilon_{ijk}e_k = e_i \times e_j = \phi(\sigma_i) \times \phi(\sigma_j) $$ where $\epsilon_{ijk}$ is the Levi-Civita symbol and $\times$ denotes the usual cross product. Hence $$ \phi([A, B]) = \phi(A) \times \phi(B),\qquad A,B \in \su $$ by linearity which implies that $\phi$ is an isomorphism of algebras of $\su$ with $(\mathbf R^3, \times)$. Now let $A, B \in \su$ be linearly independent, then $\phi(A)$ and $\phi(B)$ span a two-dimensional linear subspace $V$ of $\mathbf R^3$, hence $\phi(A) \times \phi(B)$ is a non-zero vector orthogonal to $V$; that is, $\phi([A,B]) \not\in V$, giving - as $\phi$ is invertible - $[A,B]\not\in \phi^{-1}[V] = \<A,B>$. So $A, B$ do not generate a two-dimensional subalgebra. As $A$ and $B$ were arbitrary, there isn't any.
Therefore $\su$ and $\sl$ aren't isomorphic.