Two-sign representations of the Poincare group
Notice that on the space of physical states there is realized projective representations of the Poincare group.
Namely, for acting of two unitary operators of Poincare group we can write down that
$$
\tag 1 U(\Lambda_{1})U(\Lambda_{2}) = e^{i\varphi (\Lambda_{1}, \Lambda_{2})}U(\Lambda_{1}\Lambda_{2})
$$
where $\varphi (\Lambda_{1}, \Lambda_{2})$ is called the projective phase.
There is the question: is it possible to redefine generators $P_{\mu}, J_{\mu \nu}$ (the latter represents algebra of the Lorentz group) of Poincare group so that the phase will disappear from commutation relations.
There is theorem which states that projective representations can be reduced to usual ones if
1) The group is simply connected;
2) There is possible only such solutions for phases $\varphi$ for which they can be completely adsorbed by generators redefinition (phases have to satisfy associativity conditions for operators, and Bianchi identity for generators).
Poincare group satisfies the second condition, but there is the problem with the first one. Really, it can be shown that the Lorentz group (its proper subgroup) is just
$$
\text{SO}^{\uparrow}(3,1) \simeq \text{SL}(2,C)/Z_{2}
$$
Next, the topology of $\text{SL}(2,C)$ is equal to the one of $SU(2)\times SU(2)$, which is $S_{3}\times S_{3}$. First homotopy group of $S_{3}\times S_{3}/Z_{2}$ isn't zero, and the group is doubly connected. By the close analogy, the Poincare group has topology of $R^{4}\times S_{3}\times S_{3}/Z_{2}$ (you've just have to add abelian translation group space $R^{4}$), which is again doubly connected.
This leads to the statement that the phase can be only $\pm 1$. So $(1)$ takes the form
$$
\tag 2 U(\Lambda_{1})U(\Lambda_{2}) = \pm U(\Lambda_{1}\Lambda_{2}),
$$
and we have the statement that there is realization of two-sign Poincare (Lorentz) group representations on physical states.
To avoid taking care of $\pm$ sign, we may just extend the Lorentz group to $\text{SL}(2,C)$, as you've mentioned.
Your special question
The previous item of my answer (The Lorentz group and the spin) is necessary,
So, lets answer on your question by using previously discussed things. As ACuriousMind mentioned in the comment section, there is no isomorphism
$$
\text{SO}(3,1)\simeq \text{SU}(2)\times \text{SU}(2)
$$
There is, however, isomorphism
$$
\text{SO}(3,1) \simeq \text{SL}(2,C)/Z_{2} \simeq \text{SU}(2)\times \text{SU}(2)/Z_{2}
$$
Let me describe the simple way how can you get the second equality, namely,
$$
\text{SO}(3,1) \simeq \text{SU}(2)\times \text{SU}(2)/Z_{2},
$$
since this will answer on your question.
You've started from generators of the Lorentz group, the boosts, $K_{i}$, and rotations, $R_{i}$. They satisfy commutation relations
$$
[R_{i},R_{j}] = i\epsilon_{ijk}R_{k},\quad [K_{i},K_{j}] = i\epsilon_{ijk}R_{k},\quad [K_{i},R_{j}] = -i\epsilon_{ijk}K_{k}
$$
Let's introduce new operators,
$$
\tag 3 P^{\pm}_{i} =\frac{1}{2}(R_{i}\pm K_{i})
$$
They form the algebra
$$
[P^{\pm}_{i}, P^{\pm}_{j}] = i\epsilon_{ijk}P_{k}, \quad [P_{i}^{\pm},P^{\mp}_{j}] = 0
$$
So $P^{+}, P^{-}$ generators satisfy $SU(2)$ group algebra. This, however, doesn't mean that the Lorentz group is isomorphic to $SU(2)\times SU(2)$. This means, in particular, that only the part of $SU(2)\times SU(2)$ representations are ones of $SO(3,1)$. Only by using the argument from the first section of my answer (there are realized two-sign representations of the Lorentz group on the physical states) you may use the irreps of the group $\text{SU}(2)\times \text{SU}(2)$ for classifying the physical states.
Irreps of $SU(2)$ are labelled by integer or half-integer label $\sigma$, and it has dimension $2\sigma + 1$, i.e., they the vectors
$$
|s\rangle = \begin{pmatrix}s_{1}\\... \\s_{2\sigma+1}\end{pmatrix} : \quad \hat{C}|s\rangle = -\sigma(\sigma +1)|s\rangle ,
$$
where $C$ is Casimir operator for $SU(2)$ group.
So irreps of $SU(2)\times SU(2)$ group can be labelled as $(\sigma_{1},\sigma_{2})$. Finally, from the definition you have that
$$
\frac{1}{2}(P^{+}_{i} +P^{-}_{i}) = R_{i},
$$
where $R_{i}$ is rotation generator (and associated quantity is, of course, the spin), and since
$$
\sum_{i}R_{i}^{2} = C^{2}= -S(S+1) = (\frac{1}{2}(P^{+}_{i} +P^{-}_{i}))^{2},
$$
you have that
$$
C^{2} = -(\sigma_{1}+\sigma_{2}+1)(\sigma_{1}+\sigma_{2}) = S(S+1)
$$
From this equality you have that the spin of $(\sigma_{1},\sigma_{2})$ is $S = \sigma_{1}+\sigma_{2}$.
