Fact 1: If $\mathfrak g$ is a real Lie algebra, then the complex representations of $\mathfrak g$ are in $1$-to-$1$-correspondence with the complex representations of its complexification $\mathfrak g_\mathbb C := \mathfrak g \otimes_\mathbb R \mathbb C$.
See e.g. Bijection between the complex representations of a real Lie algebra and the complex representations of its complexification. Also compare https://math.stackexchange.com/a/3258221/96384 for what this correspondence means, in particular the somewhat surprising consequence that mutually non-isomorphic $\mathfrak g$ can, to a large extent, have "the same" representations, namely, if they have the same complexification. E.g. $\mathfrak{sl}_2(\mathbb R)$ and $\mathfrak{su}_2$, the two real forms of their common complexification $\mathfrak{sl}_2(\mathbb C)$; or $\mathfrak{sl}_3(\mathbb R), \mathfrak{su}_3$ and $\mathfrak{su}_{1,2}$, the three real forms of their common complexification $\mathfrak{sl}_3(\mathbb C)$. [My answer in the second link emphasizes that this correspondence does not respect conjugation among the representations, but for now let's marvel at how much it does respect.]
Fact 2: Viewing a complex Lie algebra $L$ (of complex dimension $n$, say) as a real Lie algebra (of dimension $2n$) is a process called "scalar restriction", and for sake of clarity should sometimes be denoted with a functor notation, e.g. $R_{\mathbb C \vert \mathbb R} (L)$. It is of utmost importance to be aware that:
2A) This scalar restriction is not an inverse to scalar extension (complexification) which we saw in Fact 1. Rather, for a large class of Lie algebras including the semisimple ones, we have that
$$(R_{\mathbb C \vert \mathbb R}(L))_\mathbb C \simeq L \oplus L$$
i.e. restriction of a complex Lie algebra followed by extension gives out a sum of two copies of the complex Lie algebra we started with.
(But in general, the other way around will not do that. In fact, e.g. for $\mathfrak g := \mathfrak{sl}_2(\mathbb R)$, the real Lie algebra $$R_{\mathbb C \vert \mathbb R} (\mathfrak g_\mathbb C)$$ is not isomorphic to $\mathfrak g$ itself, nor to any direct sum of copies thereof.)
2B) To find, for a given complex Lie algebra $L$, a real Lie algebra $\mathfrak g$ such that $\mathfrak g_\mathbb C \simeq L$ (i.e. finding "an actual inverse" of complexification) is a harder task. Such a $\mathfrak g$ is called a "real form" of $L$. Not all complex Lie algebras have real forms. All semisimple complex Lie algebras do have real forms, but unless the Lie algebra is $0$, they are never unique. (At the end of fact 1 I listed all real forms of $\mathfrak{sl}_n(\mathbb C)$ for $n=2,3$. Bigger complex Lie algebras tend to have even more mutually non-isomorphic real forms.)
Cf. https://math.stackexchange.com/a/3975561/96384 and https://math.stackexchange.com/a/3895802/96384.
Now to apply this to your problem. By a funny coincidence, indeed the Lorentz Lie algebra $\mathfrak{so}(1,3)$, a $6$-dimensional real Lie algebra, is isomorphic to $R_{\mathbb C\vert \mathbb R}(\mathfrak{sl}_2(\mathbb C))$, the scalar restriction of $\mathfrak{sl}_2(\mathbb C)$. Consequently we have $1$-to-$1$-correspondences
complex reps of $\mathfrak{so}(1,3)$ $\leftrightarrow$ complex reps of $R_{\mathbb C\vert \mathbb R}(\mathfrak{sl}_2(\mathbb C))$ $\stackrel{\text{Fact }1}\leftrightarrow$ complex reps of $(R_{\mathbb C\vert \mathbb R}(\mathfrak{sl}_2(\mathbb C)))_\mathbb C$ $\stackrel{\text{Fact }2A}\leftrightarrow$ complex reps of $\mathfrak{sl}_2(\mathbb C) \oplus \mathfrak{sl}_2(\mathbb C)$
So you see, via the "doubling" process in 2A, the representations of $\mathfrak{sl}_2(\mathbb C)$ viewed as real Lie algebra are actually in correspondence with the the respresentations of the complex Lie algebra $\mathfrak{sl}_2(\mathbb C) \oplus \mathfrak{sl}_2(\mathbb C)$. (And similarly for any semisimple complex Lie algebra $L$, the complex reps of $R_{\mathbb C\vert \mathbb R}(L)$ are the complex reps of $L \oplus L$.)
Now one can take this further:
Fact 3: If $L_1$ and $L_2$ are two semisimple complex Lie algebras, then every irreducible representation of $L_1 \oplus L_2$ is a tensor product of an irrep of of $L_1$ with an irrep of $L_2$; and conversely, each such tensor product of irreps of the factors gives an irrep of the direct sum.
See e.g. Tensor product of simple modules over semisimple Lie algebras, or Tensor product of irreducible representations of semisimple Lie algebras. There are tons of questions and answers on this site for the corresponding case of groups (where of course the direct sum is replaced by a direct product).
Applying this to your problem we see that for irreducible reps, we can prolong the above chain, and identify
irreducible complex reps of $\mathfrak{so}(1,3) \leftrightarrow ... \leftrightarrow$ irreducible complex reps of $\mathfrak{sl}_2(\mathbb C) \oplus \mathfrak{sl}_2(\mathbb C) \stackrel{\text{Fact }3} \leftrightarrow$ tensor products of two irreducible complex reps of $\mathfrak{sl}_2(\mathbb C) \stackrel{\text{Fact }1} \leftrightarrow$ tensor products of two irreducible complex reps of $\mathfrak{su}_2$
where in the last step we just made use of the fact that $\mathfrak{su}_2$ is a real form of $\mathfrak{sl}_2(\mathbb C)$. Note that since $\mathfrak{sl}_2(\mathbb R)$ is another real form of $\mathfrak{sl}_2(\mathbb C)$, we could also have identified this with the tensor products of two irreducible complex reps of $\mathfrak{sl}_2(\mathbb R)$.
Exercise: Show that the complex irreps of $\mathfrak{sl}_{\color{red}{3}}(\mathbb C)$ viewed as a real Lie algebras are in $1$-to$1$-correspondence with (many things, e.g.) the tensor products of two irreps of $\mathfrak{su}_{1,2}$.
