I know that given an irreducible $\mathfrak sl_2$ module $V$, it is $V\cong V(m), m\in \mathbb Z_+$, whose basis $(v_0,\cdots,v_m$) is such that:
$$H\cdot v_j = (m-2j)v_j, Y\cdot v_j = (j+1)v_{j+1}, X\cdot v_j = (m-j+1)v_{j-1}, \mbox{for } j\geq 0 \mbox{ and }v_{-1} = 0 $$
and $V(m) = \bigoplus^m_{k= -m} V_{m-2k}$.
Now given $V=V(m), W=W(n)$ irreducible $\mathfrak sl_2$ modules, why is that $V\otimes W \cong V({m+n})\oplus \cdots \oplus V({|n-m|})$?
Well, we can choose such a basis for $V$ and $W$, respectively: $(v_0,\cdots, v_m)$ and $(w_0,\cdots, w_n$). Then the set $v_i\otimes w_j, 0\leq i \leq n, 0 \leq j \leq m ,$ is a basis for the tensor product $V\otimes W$. Ok, now applying an element of the cartan subalgebra give us:
$H\cdot v_i\otimes w_j = (m+n-2(i+j))v_i\otimes w_j.$
Since $i+j$ lies in the set $\{1,2,\cdots, n+m\}$, it follows that the eigenvalue for $v_i\otimes w_j$ also lies in the set $\{n+m, n+m-2, \cdots, -n-m+2,-n-m\}$. Now how to proceed from here? Looking for what we got to prove, we see that there are no negative eigenvalue anymore; why is that? any hint on how to proceed?
Define the character $\text{ch}_V$ of a (finite-dimensional) representation of $\frak sl_2$ as $\sum_k \dim E_k(V) t^k$ where $E_k(V)$ is the eigenspace of $H$ acting on $V$ for the eigenvalue $k$. Then $\text{ch}(V\oplus W) =\text{ch}(V)+\text{ch}(W)$ and $\text{ch}(V\otimes W) =\text{ch}(V)\text{ch}(W)$. Also $$\text{ch}(V(n))=t^n+t^{n-2}+t^{n-4}+\cdots +t^{-n}=\frac{t^{n+1}-t^{-n-1}} {t-t^{-1}}.$$ The characters for the $V(n)$ are linearly independent, and the character determines a representation up to isomorphism.
Let $n\ge m$. Then \begin{align} (t-t^{-1})\text{ch}(V(n))\text{ch}(V(m))&=(t^{n+1}-t^{-n-1}) (t^m+t^{m-2}+\cdots+t^{-m})\\ &=(t^{m+n+1}-t^{-m-n-1})+(t^{m+n-1}-t^{-m-n+1})+\cdots +(t^{m-n+1}-t^{-m+n-1})\\ &=(t-t^{-1})(\text{ch}(V(m+n))+\text{ch}(V(m+n-2))+\cdots+\text{ch}(V(m-n))) \end{align} and so $$V(n)\otimes V(n)\cong V(m+n)\oplus V(m+n-2)\oplus\cdots\oplus V(m-n).$$