Setting spinors and $SU(2)$ representations on the same patch I am sorry for the naivety of this question, I am a mathematician and I am trying to put together different ideas.
I am trying to understand the vocabulary of physics, in particular, I want to know:
We know that the isometry group of Minknowski metric is the Lorentz group $O(3,1)$. Furthermore, it acts on $\mathbb R^{3,1}$ by isometries. The isotropy group of this action is precisely $SU(2)$, so it is interested to understand further representations of $SU(2)$.
$1)$ What does mean the notation $(1/2,0)\oplus (0,1/2)?$
$2)$ How can I express an spinor from the $1/2$-representation of $SU(2)$?
 A: Reference: Srednicki, Quantum Field Theory ch. 33
The Lie algebra of the Lorentz group $SO(3,1)$ can be described in terms of 3 spacial rotations ($J_i$) and 3 velocity boosts ($K_i$):
$$ [J_i,J_j] = i \epsilon_{ijk} J_k $$
$$ [J_i,K_j] = i \epsilon_{ijk} K_k $$
$$ [K_i,K_j] = -i \epsilon_{ijk} J_k $$
To make the $SU(2) \otimes SU(2)$ structure of the (complexified) algebra explicit we need to change to a different set of generators:
$$ N_i = \frac{1}{2} (J_i - iK_i) $$
$$ N_i^\dagger = \frac{1}{2} (J_i + iK_i) $$
This gives:
$$ [N_i,N_j] = i \epsilon_{ijk} N_k $$
$$ [N_i^\dagger,N_j^\dagger] = i \epsilon_{ijk} N_k^\dagger $$
$$ [N_i,N_j^\dagger] = 0 $$
So given that the irredicible representations of $SU(2)$ can be listed by a single index $j= 0, \frac{1}{2} , 1, \frac{3}{2} ...$ , irreducible representations of $SU(2) \otimes SU(2)$ clearly needs two such indices $(j_1,j_2)$ where $j_1$ and $j_2$ are the eigenvalues of $N_3$ and $N_3^\dagger$ above.
Important representations in physics are:
$(0,\frac{1}{2})  $ = Weyl Spinor
$(\frac{1}{2},\frac{1}{2})  $ = Vector
$(0,\frac{1}{2}) \oplus (\frac{1}{2},0)  $ = Dirac Spinor
For each of these representations in turn we can calculate the corresponding SU(2) representations of the isotropy group (physicists usually call this the little group). 
Physically the little group (isotropy group) corresponds to the subgroup of the Lorentz group which leave the momentum vector unchanged. This (at least for a massive particle) is the group of spatial rotations i.e. the $J_i$ above.
So we need to write $J_i$ in terms of the $N_i$ and $N_i^\dagger$ :
$$ J_i = N_i + N_i^\dagger $$
All physicists then recognise this as the process of "adding two angular momentum" (mathematically it is the process of finding the irreducible representations of the tensor product of two representations):
$(j_1,j_2)$ decomposes into the direct sum of representations $|j_1-j_2|, |j_1-j_2| + 1 , ... j_1 + j_2 $
Applying this we get:
$(0,\frac{1}{2}) :  j = \frac{1}{2}$  so this is a Spin $\frac{1}{2}$ particle.
$(\frac{1}{2},\frac{1}{2}) :  j = 0,1$  so this contains a Spin 1 (vector) particle.
$(0,\frac{1}{2}) \oplus (\frac{1}{2},0) :  j = \frac{1}{2} $ in both parts of the sum, so this again is a Spin $\frac{1}{2}$ particle. 
