Let $\mathbf{x} = (x_{1}, x_{2}, x_{3})^T$ be the coordinates of a point in the $e$-basis and let $\mathbf{y} = (y_{1}, y_{2}, y_{3})^T$ be the coordinates of the same point in the $f$-basis.
It is the same point, so we require the following condition.
$$
x_{1} \mathbf{e}_{1} +
x_{2} \mathbf{e}_{2} +
x_{3} \mathbf{e}_{3}
=
y_{1} \mathbf{f}_{1} +
y_{2} \mathbf{f}_{2} +
y_{3} \mathbf{f}_{3}
$$
The question gives the way of writing the $f$-basis vectors in terms of the $e$-basis vectors:
$$
\begin{aligned}
\mathbf{f}_1 &= \mathbf{e}_1 + \mathbf{e}_2 \\
\mathbf{f}_2 &= \mathbf{e}_2 \\
\mathbf{f}_3 &= \mathbf{e}_1 - \mathbf{e}_3
\end{aligned}
$$
We can substutite these formulas into the equation for the coordinates above
$$
x_{1} \mathbf{e}_{1} +
x_{2} \mathbf{e}_{2} +
x_{3} \mathbf{e}_{3}
=
y_{1} (\mathbf{e}_{1} + \mathbf{e}_{2}) +
y_{2} \mathbf{e}_{2} +
y_{3} (\mathbf{e}_{1} - \mathbf{e}_{3})
$$
$$
x_{1} \mathbf{e}_{1} +
x_{2} \mathbf{e}_{2} +
x_{3} \mathbf{e}_{3}
=
y_{1} \mathbf{e}_{1} + y_{1} \mathbf{e}_{2} +
y_{2} \mathbf{e}_{2} +
y_{3} \mathbf{e}_{1} - y_{3} \mathbf{e}_{3}
$$
$$
x_{1} \mathbf{e}_{1} +
x_{2} \mathbf{e}_{2} +
x_{3} \mathbf{e}_{3}
=
(y_{1} + y_{3}) \mathbf{e}_{1}
+
(y_{1} + y_{2} ) \mathbf{e}_{2}
- y_{3} \mathbf{e}_{3}
$$
Now $\mathbf{e}_1$, $\mathbf{e}_2$ and $\mathbf{e}_3$ are three linearly independent vectors so that the components of each vector can be equated on both sides of the above. I.e. we can write:
$$
\begin{aligned}
x_1 &= y_1 + y_3 \\
x_2 & = y_1 + y_2 \\
x_3 &= -y_3
\end{aligned}
$$
The following is exactly the same as the above with slightly different spacing
$$
\begin{aligned}
x_1 &= y_1 & &+y_3 \\
x_2 & = y_1 &+ y_2 & \\
x_3 &= & &-y_3
\end{aligned}
$$
Writing the equations that relate the coordinates in this way, we can see how the set can be written as a single matrix equation:
$$
\begin{pmatrix}
x_{1} \\ x_{2} \\ x_{3}
\end{pmatrix}
=
\begin{pmatrix}
1 & 0 & 1 \\
1 & 1 & 0 \\
0 & 0 & -1
\end{pmatrix}
\begin{pmatrix}
y_{1} \\ y_{2} \\ y_{3}
\end{pmatrix}
\Rightarrow
\mathbf{x}
=
\begin{pmatrix}
1 & 0 & 1 \\
1 & 1 & 0 \\
0 & 0 & -1
\end{pmatrix}
\mathbf{y}$$
This shows how we can use a matrix to convert coordinates in the $f$-basis to coordinates in the $e$-basis, i.e. $\mathbf{x} = T \mathbf{y}$, i.e. it represents the matrix $T$ in the formula
$$
A' = T^{-1} A T
$$
where $A$ is the transformation that is applied to coordinates in the $e$-basis. The above formula is applied
Having found $T$, we can find its inverse (by hand or with some software):
$$
T^{-1}
=
\begin{pmatrix}
1& -1& 0 \\
0& 1& 0 \\
1& -1& -1
\end{pmatrix}
$$
and finally, we can calculate $A'$
$$
A'
=
\begin{pmatrix}
-3& -2& -2 \\
3& 3& -1 \\
-7& -1& -6
\end{pmatrix}
$$
which is the matrix of the transformation that is applied to coordinates in the $f$-basis.
This next part goes into how the formula relating the two transformation matrices in the different bases is derived.
If we write $\mathbf{u}$
for the result of applying $A$ to the $e$-basis vector $\mathbf{x}$ and if we write
$\mathbf{v}$ for the result of applying $A'$ to the $f$-basis vector $\mathbf{y}$.
$$
\begin{aligned}
\mathbf{u} &= A \mathbf{x} \\
\mathbf{v} &= A' \mathbf{y} \\
\end{aligned}
$$
The vectors
$\mathbf{x}$ and $\mathbf{y}$
correspond to the same point in the two different bases and so do the pair of vectors
$\mathbf{u}$ and $\mathbf{v}$. In other words they can be written:
$$
\begin{aligned}
\mathbf{x} &= T \mathbf{y} \\
\mathbf{u} &= T \mathbf{v} \\
\end{aligned}
$$
This means we can write the following
$$
\begin{aligned}
\mathbf{u} &= T \mathbf{v} \\
A \mathbf{x} &= T \mathbf{v} \\
A \mathbf{x} &= T A' \mathbf{y} \\
A T\mathbf{y} &= T A' \mathbf{y} \\
T^{-1} A T\mathbf{y} &= A' \mathbf{y} \\
\end{aligned}
$$
As this works for all $\mathbf{y}$ we can conclude that $T^{-1} A T = A' $.
Best Answer
Suppose that you have a closed box in Sweden which you wish to be open in Sweden. You only know how to open the box if you are in China. Then
(open box in Sweden) = (go from Sweden to China) THEN (open box in China) THEN (go from China to Sweden)
(note that function composition is written funny in math compared to the English word "THEN.")