How to formulate with a tensor multiplication method $2\otimes 2\otimes 2$ in $SU(2)$ group? In $SU(2)$ we can write $2\otimes 2=3\oplus 1$ or 
\begin{equation}
q_iq^j=\left(q_iq^j-\frac{1}{2}\delta^i_jq^kq_k\right)+\frac{1}{2}\delta^i_jq^kq_k,
\end{equation}
where $q_i$ is a $SU(2)$ doublet, the first term of the rhs transforms as a triplet and the second one as a singlet.
With this representation, we can express triplets depending on the $q_i$ content. For instance, if $q=\left(u\;d\right)^T$, then a pion triplet is obtained.
I'm looking for a $SU(2)$ quartet that is supposed to be obtained from $2\otimes 2\otimes 2=2\oplus 2\oplus 4$. I mean, I would like to find a tensor equation like the one written above.
 A: The Young Tableau of the tensor product will tell you the symmetries of the tensors that transform in each representation. Since we don't have access to any Young Tableaux packages I will have to draw them with ASCII.
For example, $2 \otimes 2 = 3 \oplus 1$ looks like
$$▊\otimes▊ = ▊▊ + \cdot,$$
where $\cdot$ means the trivial, one-dimensional rep. The $3$ is represented by $▊▊$ and boxes in the same row represent the process of symmetrising indices and subtracting the trace. This is what OP wrote in their first equation. The trivial rep is the trace, $q^{k}q_{k}$, which OP has written with the invariant $\delta^{i}{}_{j}$. 
Let's continue for $2 \otimes 2 \otimes 2 = (3 \oplus 1)\otimes 2 = (3 \otimes 2) \oplus (1 \otimes 2) = [4 \oplus 2] \oplus 2 = 4 \oplus 2 \oplus 2$. In Young Tableux we get
$$▊\otimes ▊ \otimes ▊ = (▊▊ \oplus \cdot)\otimes ▊ = (▊▊\otimes▊)\oplus (▊ \otimes \cdot) = [▊▊▊ \oplus ▊] \oplus ▊ = ▊▊▊ \oplus ▊ \oplus ▊$$
where I have used that 
$$ \begin{matrix}▊ &\!\!\!\!\! ▊ \\ ▊\end{matrix} = ~ ▊$$ because two boxes in the same column are antisymmetrised away (for $SU(N)$ it would be $N$ boxes).   
The $4$ representation corresponds to $▊▊▊$ which is fully symmetrised in the three indices and with the trace subtracted. 
Does that help you to write down the tensor that transforms in that representation?
A: You might profit from the Introduction to unitary symmetry book by P Carruthers in learning the index language of tensoring SU(N) representations, but it is obvious from your comments that this would be runaway overkill, as you just want quark wavefunctions of hadrons, gotten far more directly and, in the case of SU(2), same as in undergraduate QM spin laddering, trivially.  (Your pion example is doubly fraught, since it involves the conjugate rep with upstairs indices, and the funny - sign that attaches to it.)
For spin, recall from elementary QM that adding three spin doublets yields you a spin j =3/2 quartet, and two spin j =1/2 doublets. The quartet is perfectly symmetric, so its $j_z=3/2$ component is just $|\uparrow \uparrow \uparrow \rangle$, which spin-ladders ($J_-$) down to the spin $j_z=1/2$ component,
$$
\psi_{10}=\frac {1}{\sqrt{3}}(|\downarrow \uparrow \uparrow \rangle+|\uparrow \downarrow \uparrow \rangle+|\uparrow \uparrow \downarrow \rangle),
$$
perforce orthogonal to the two spin 1/2 doublets with the same spin $j_z=1/2$ components,
$$
\psi_{8}=\frac {1}{\sqrt{2}}(-|\downarrow \uparrow \uparrow \rangle+ |\uparrow \uparrow \downarrow \rangle),\qquad 
\psi_{8'}=\frac {1}{\sqrt{6}}(|\downarrow \uparrow \uparrow \rangle-2 |\uparrow \downarrow \uparrow \rangle+|\uparrow \uparrow \downarrow \rangle),
$$
which are also orthogonal to each other, and, naturally, both up-ladder to 0.
Ignore spins now, and reinterpret them as isospin, and the $\psi_{10}$ will give you  the $\Delta$ baryon (in the eightfold way decuplet, in which you might not be interested, by all appearances. You might have to multiply its wave function by a corresponding spin such, and a color-antisymmetrized one, but you also might wish to not look there yet.). So your $\Delta^+$ is but $\frac {1}{\sqrt{3}}(| duu\rangle  +| udu\rangle + | uud\rangle )$. The mixed symmetry isodoublets also combine to put together the mixed symmetry nucleon isodoublet (again, mixed-symmetrize with spin, and fully antisymmetrized by color, in the eightfold way octet). 
If that's all you wanted, you are done.
