Recall what the matrix $[\, T\,]_B$ is:
If you write a vector $x\in\Bbb R^2$ in terms of the basis $B=\{\,b_1,b_2\,\}$ $$ x=\alpha_1 b_1+\alpha_2 b_2, $$ then if you multiply $[\, T\,]_B$ by the coordinate vector $x_B=\bigl[{\alpha_1\atop\alpha_2}\bigr]$, you get the coordinate vector of $T(x)$ with respect to $B$. That is $$\tag{1} [T(x)]_B = [\,T\,]_B x_B. $$
Now the columns of the matrix $[\,T\,]_E$ where $E=\{e_1,e_2\}$ are the vectors $T(e_1)$ and $T(e_2)$. To find these vectors, we can use the matrix $[\, T\,]_B$. There are three steps involved here. Considering the vector $e_2$, we have to
- Find the coordinates of $e_2$ with respect to the basis $B$.
- Find the coordinates of $T(e_2)$ with respect to the basis $B$.
- Find $T(e_2)$ expressed in the standard basis.
Step 1: For $e_2=(0,1)$, we first find the coordinates of $e_2$ in terms of the basis $B$. Towards this end, we have to solve the system $$ \Bigl[{0\atop1}\Bigr]= \alpha_1 \Bigl[{-1\atop-3}\Bigr] +\alpha_2\Bigl[{-3\atop-10}\Bigr]. $$ Doing so gives: $$\alpha_1=3,\quad \alpha_2=-1$$ The coordinate vector of $e_2$ with respect to $B$ is $\bigl[{3\atop-1}\bigr]$.
Note we could have done this differently: the coordinate vector $\bigl[{\alpha_1\atop\alpha_2}\bigr]$ of $x$ with respect to $B$ satisfies $[b_1 \ b_2] \bigl[{\alpha_1\atop\alpha_2}\bigr] =x$; so $\bigl[{\alpha_1\atop\alpha_2} \bigr]=[b_1 \ b_2]^{-1} x$. Thus we could have found $[b_1\ b_2]^{-1}$ and just multiplied this by $e_2$. This is actually preferable, since we can use the inverse when considering $e_1$ later.
Step 2: Using $(1)$ now, the coordinate vector of $T(e_2)$ with respect to $B$ is $$ [\,T\,]_B (e_2)_B = \Bigl[ \matrix{4&4\cr 4&5 }\Bigr]\Bigl[{3\atop -1} \Bigr] = \Bigl[{8\atop 7} \Bigr]. $$
Step 3: But note that $T(e_2)$ is not the vector $\bigl[{8\atop 7} \bigr]$; this vector gives the coordinates of $T(e_2)$ with respect to the basis $B$. In general, if $x_B$ is the coordinate vector of $x$ with respect to $B$, then $x=[b_1\ b_2] x_B$; so $$ T(e_2)=[b_1\ b_2]\Bigl[{8\atop 7} \Bigr] =8\, b_1+7\, b_2= 8\Bigl[{-1\atop -3} \Bigr] +7 \Bigl[{-3\atop -10} \Bigr] =\Bigl[{-29\atop -94} \Bigr]. $$
Thus, the second column of $[\,T\,]_E$ is $\bigl[{-29\atop -94} \bigr]$.
To find the first column of $[\,T\,]_E$, apply the same procedure to the vector $e_1$. The first step here would be to write $e_1$ in terms of the basis $B$. To do that, you need to solve the system $$ \Bigl[{1\atop0}\Bigr]= \alpha_1 \Bigl[{-1\atop-3}\Bigr] +\alpha_2\Bigl[{-3\atop-10}\Bigr]. $$ or compute: $$\Bigl[{\alpha_1\atop\alpha_2} \Bigr]=[b_1 \ b_2]^{-1} e_1.$$
In general, given $x$ written in terms of the standard basis, the coordinates of $x$ with respect to
$B$ are given by $ P^{-1} x$ where $P=[b_1\ b_2]$. Then the coordinates of $T(x)$ with respect to
$B$ are $[\,T\,]_B P^{-1} x$. Then we have that
$T(x)$ expressed in terms of the usual basis is
$P [\,T\,]_B P^{-1} x$. So, $$ [\,T\,]_E = P[\,T\,]_B P^{-1}. $$