Composition of linear transformations verification 

I have some difficulties understanding how the change of Basis works on polynomial.
Is $(k \circ l )_{B \leftarrow B}$ = $(k \circ l )_{B \leftarrow  c}$.$(k \circ l )_{B \leftarrow c}$  which is the same as $(k \circ l )_{B}$ .$(k \circ l )_{B^-1}$ ?

Is that the composition of k and l?

$(k \circ l_1 )= -6m_1 +6m_0$
$(k \circ l_2)=-5m_1+ 1m_0$

 A: First off, none of what you wrote is a change of basis.
Linear Transformations and Transformation Matrices
Recall that a linear transformation $l: V \rightarrow W$ between finite dimensional $k$-vector spaces is determined on a basis $\{v_1, \ldots, v_m\}=:B$ of $V$. This means that to give $l$, it suffices to know what $l(v_i) \in W$ is for every $i \in \{1, \ldots, m\}$. 
Since $l(v_i) \in W$ it can be given as a linear combination of a basis $\{w_1, \ldots, w_n\}=:C$ of $W$. I.e. for every $i \in \{1, \ldots, m\}: l(v_i) = \sum_{j = 1}^n a_{ij} w_j$. Given the $v_i$ and the $w_j$, the linear transformation is thus uniquely determined by the $a_{ij} \in k$. These $a_{ij}$ determine the transformation matrix of $l$ with respect to the bases $B$ and $C$. Write $(l)_{C \leftarrow B}$ (other notations are common too) and call it the transformation matrix* of $l$ with respect to the bases $B$ and $C$. In your first example this means that 
$$l(m_1) = \underbrace{3}_{=a_{11}} e_1 + \underbrace{(-3)}_{=a_{12}} e_2 ~~~ \text{and} ~~~ l(m_2) = \underbrace{2}_{=a_{21}} e_1 + \underbrace{(-1)}_{=a_{22}} e_2 $$
which gives 
$$ l_{C \leftarrow B} = \begin{pmatrix} 3 & 2 \\ -3 & -1 \end{pmatrix}$$
Note that the transformation matrix of a linear transformation are formed with respect to a basis $B$ of $V$ and a basis $W$ of $W$ i.e. after a choice of bases.
Composition and Matrix Multiplication
For the transformation matrix of the composition of two linear transformations consider the following: 
Let $f:A \rightarrow B$, $g: B \rightarrow C$ be linear transformations between $k$ vector spaces with bases $\{a_1, \ldots, a_m\}=: A'$, $\{b_1, \ldots, b_n\}=: B'$, $\{c_1, \ldots, c_p\}=: C'$ respectively. Then consider again what the composition does to a basis $\{a_1, \ldots, a_m\}$ of $A$: 
$$\begin{align} (g \circ f)(a_i) &= \sum_{k=1}^p u_{ik} c_k \end{align}$$
On the other hand with $g(b_j) = \sum_{k=1}^p t_{jk} c_k$ for every $j \in \{1, \ldots, n\}$
$$\begin{align} g (f(a_i)) &= g(\sum_{j=1}^n s_{ij} b_j) \\
&= \sum_{j=1}^n s_{ij} g(b_j) \\
&= \sum_{j=1}^n s_{ij} \sum_{k=1}^p t_{jk} c_k \\
&= \sum_{k=1}^p \sum_{j=1}^n s_{ij} t_{jk} c_k 
\end{align}$$
Now note that 
$$ \sum_{j=1}^n s_{ij} t_{jk} = u_{ik}$$
Here the $u_{ik}$ are the entires of the transformation matrix $(g \circ f)_{C' \leftarrow A'}$. Hence we have 
$$ (g \circ f)_{C' \leftarrow A'} = (g)_{C' \leftarrow B'} \cdot (f)_{B' \leftarrow A'}$$ with the usual matrix multiplication. 
For your problem this means that you'll first need to determine the transformation matrix $(k)_{B \leftarrow C}$, given by the entries $b_{ij}$. As suggested in the first paragraph, this is done by determining the action of $k$ on the basis vectors. In your case $e_1$ and $e_2$: 
$$k(e_1) = k(\begin{pmatrix}1 \\ 0 \end{pmatrix}) = \underbrace{(-3)}_{=b_{11}} m_1 + \underbrace{(-1)}_{=b_{12}} m_2$$ 
and 
$$k(e_2) = k(\begin{pmatrix}0 \\ 1 \end{pmatrix}) = \underbrace{(-1)}_{=b_{21}} m_1 + \underbrace{(-3)}_{=b_{22}} m_2 $$
Your transformation matrix is thus 
$$ (k)_{B \leftarrow C} = \begin{pmatrix} -3 & -1 \\ -1 & -3 \end{pmatrix}$$
Finally, the transformation matrix of the composition is given by the matrix product of the two: 
$$ (k\circ l)_{B \leftarrow B} = k_{B \leftarrow C} \cdot l_{C \leftarrow B} = \begin{pmatrix} -3 & -1 \\ -1 & -3 \end{pmatrix} \cdot \begin{pmatrix} 3 & 2 \\ -3 & -1 \end{pmatrix} = \begin{pmatrix} -6 & -5 \\ 6 & 1 \end{pmatrix}$$
