Why are the basis of the space of modular form in $SL_2(\mathbb{Z})$ of weight 12 ($M_{12}(SL_2(\mathbb{Z})$)
$E_3^4$ and $\Delta$?
There is a theorem that says that the generators of modular form of weight $k$ are $E_4^{\alpha} \cdot E_6^{\beta}$ with $4\alpha +6\beta=k$. So shouldn't the basis be $E_4^3$ and $E_6^2$??
I don't know how the $\Delta$ appears. All I know is that $\Delta$ is of weight 12 and it has integral coefficients in the q-expansion just like $E_4$ and $E_6$
I can't seem to grasp this basic idea. Does it have to do with
$M_k = S_k + \mathbb{C}G_k$? ( the plus '+' is actually the tensor product.)
Yes, the Theorem is correct: a basis of $\mathcal{M}_k$ are all monomials $E_4^aE_6^b$ with $4a+6b=k$, and $a,b\ge 0$ integers. With $k=12=4a+6b$ we have either $(a,b)=(3,0)$ or $(a,b)=(0,2)$, because $a,b\ge0$ are integers. So the dimension is $2$. Of course, the Theorem gives, in general that the space $\mathcal{M}_k$ has dimension $$ \dim \mathcal{M}_k=\begin{cases} [k/12] & \text{ if } k\equiv 2\bmod 12\\ [k/12]+1 & \text{ if } k\not\equiv 2\bmod 12\end{cases} $$ Since $k=12$ is not congruent to $2$ modulo $12$ we obtain dimension $1+1=2$.
As to $\Delta$, there is the subspace $S_k\subseteq \mathcal{M}_k$ of cusp forms. For $k=12$, $\dim S_{12}=1$ with basis $\Delta$.