Suppose that we have a Lie algebra $\mathfrak{g}$ of a Lie group G, what can I say about the latter if I can find two Lie subalgefbras $\mathfrak{a},\mathfrak{b}\subset\mathfrak{g}$?
I'm asking this question because in my notes that's written that the Lie algebra of the proper orthocronus Lorentz group $\mathcal{L}_{+}^{\uparrow}$ can be decomposed into $$\mathfrak{so}(1,3)=\mathfrak{su}(2)\oplus\mathfrak{su}(2)$$ so that we conclude $$SO(1,3) ≃ SU(2)\otimes SU(2)$$ How do we reach that conclusion?
As Moishe says in the comments, both of the stated isomorphisms are incorrect. I will assume that the real question is in the title rather than the body: what can we say if a Lie algebra $\mathfrak{g}$ is the direct sum (of Lie algebras) of two subalgebras $\mathfrak{a}, \mathfrak{b}$?
What we can conclude is that if $G, A, B$ are the unique simply-connected Lie groups with Lie algebras $\mathfrak{g}, \mathfrak{a}, \mathfrak{b}$ then
$$G \cong A \times B.$$
A correct example of this is that $\mathfrak{so}(4)$ decomposes into a direct sum $\mathfrak{su}(2) \oplus \mathfrak{su}(2)$, which gives an exceptional isomorphism
$$Spin(4) \cong SU(2) \times SU(2).$$
However $\mathfrak{so}(1, 3)$ is simple, so it is not the direct sum of two of its subalgebras. It does have an exceptional isomorphism to $\mathfrak{sl}_2(\mathbb{C})$, meaning it can be written as a direct sum of vector spaces $\mathfrak{su}(2) \oplus i \mathfrak{su}(2)$, since $\mathfrak{sl}_2(\mathbb{C})$ is the complexification of $\mathfrak{su}(2)$. However, $i \mathfrak{su}(2)$ is not a subalgebra, so this is not a direct sum of Lie algebras.