Proof that $(1/2,1/2)$ Lorentz group representation is a 4-vector Taken from Quantum Field Theory in a Nutshell by Zee, problem II.3.1:
Show by explicit computation that $(\frac{1}{2},\frac{1}{2})$ is indeed the Lorentz vector.
This has been asked here:
How do I construct the $SU(2)$ representation of the Lorentz Group using $SU(2)\times SU(2)\sim SO(3,1)$ ?
but I can't really digest the formality of this answer with only a little knowledge of groups and representations.
By playing around with the Lorentz group generators it is possible to find the basis $J_{\pm i}$ that separately have the Lie algebra of $SU(2)$, and thus can be separately given spin representations.
My approach has been to write $$J_{+i}=\frac{1}{2}(J_{i}+iK_{i})=\frac{1}{2}\sigma_{i}$$ $$J_{-i}=\frac{1}{2}(J_{i}-iK_{i})=\frac{1}{2}\sigma_{i}$$ which implies that $$J_{i}=\sigma_{i}$$ $$K_{i}=0$$ However I don't really get where to go next. 
 A: Sorry this reply has come so long after your post. As this is a homework-type problem I won't perform all the calculations, but will spell out all the key points; it's probably simpler to follow this way anyway if you fill in the matrix algebra yourself!
To understand what's going on here, first of all think about a spin-1 system. In that case the action of $J_3$ on vectors is represented by the matrix $$\left(\begin{array}{ccc}
0 & -i & 0 \\
i & 0 & 0 \\
0 & 0 & 0 \end{array}  \right)$$
If you follow the standard construction of the spin-1 representation using ladder operators found in any QM textbook, you find instead that $J_3$ is represented by the matrix $$\left(\begin{array}{ccc}
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & -1 \end{array}  \right)$$
because it's acting on a basis of $J_3$ eigenstates. You can see that these matrices are unitarily equivalent by finding the eigenvalues of the first one  to be $-1,0,1$; working out the corresponding eigenvectors gives you the matrix that changes basis (up to some ordering of that basis). (It's nice to notice that this matrix transforms linearly polarised transverse vector fields into circularly polarised ones!)
Now, in the given case where we have $(1/2,1/2) = SU(2) \otimes SU(2)$. $J_3$ acts on both the (1/2,0) and (0,1/2) representations by $\frac{1}{2} \sigma_3$. Because it's a generator of the group, and not a group element, the action on the tensor product of the two groups is given by 
$$J_3 = \frac{1}{2} \sigma_3 \otimes I + I \otimes \frac{1}{2} \sigma_3 = \left(\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & -1 \end{array}  \right)$$
It's clear that this matrix gives us the action of $J_3$ on its eigenstates, and a change of basis gives us the 4-vector representation. The difference is that we now have two 0 eigenvalues; these have resulted from "adding the angular momentum" of our two spin-1/2 representations. One is from the spin-1 rep as before. The other is from the spin-0 representation; this is your time direction in the 4-vector, which is a scalar under rotations. So in this case the components of the matrix that changes basis acts on the 0-eigenvalue directions are the Clebsch-Gordon coefficients.
Hope that helps.
A: You have misunderstood the strategy. The idea is that we define new operators
$A_i = \frac{J_i + iK_i}{2}$ and $B_i =\frac{J_i-iK_i}{2} $. 
Now compute $[A_i,A_j], [B_i,B_j] $ and $[A_i,B_j]$ The answer you should get is that the first two should give you the lie algebra of $SU(2)$ while the second should give you zero. This means we have at least locally $SU(2)\times SU(2)$
The lorentz boosts for a two component spinor is then $e^{i\vec{\sigma }\cdot \theta + \vec{\sigma} \cdot \phi}$ and the other is $e^{i\vec{\sigma }\cdot \theta - \vec{\sigma} \cdot \phi}$. Direct sum of this does the general job
i.e $\begin{pmatrix}e^{i\vec{\sigma }\cdot \theta + \vec{\sigma} \cdot \phi}& 0 \\
0 &e^{i\vec{\sigma }\cdot \theta - \vec{\sigma} \cdot \phi} \end{pmatrix}$
So from this you should be able to get the generators you want
