Universal Covering Group of $SO(1,3)^{\uparrow}$

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.