I don't know whether you are dealing with $2\times 2$ matrices or general $n \times $n$ matrices. The result is true in either case.
It may not be clear to you what these spaces are. Define addition of matrices by adding corresponding entries. So for example
$$\begin{bmatrix}
1 & 2\\
0 & 3
\end{bmatrix}
+
\begin{bmatrix}
5 & 3\\
0 & 1
\end{bmatrix}
=
\begin{bmatrix}
6 & 5\\
0 & 4
\end{bmatrix}
$$
If $c$ is a constant (a scalar, a number) then you multiply a matrix by $c$ by multiplying each entry by $c$. So for example
$$3\begin{bmatrix}
1 & 2\\
0 & 3
\end{bmatrix}
=
\begin{bmatrix}
3 & 6\\
0 & 9
\end{bmatrix}
$$
A vector space of matrices is a collection $V$ of matrices (of the same size) such that if $A$ and $B$ are matrices in the collection, then so is the sum $A+B$, and also if $c$ is any scalar, then $cA$ is in the collection.
So typically a vector space of matrices will have many matrices in it. The only vector space of matrices that consists of a single matrix is the space whose only element is the all $0$'s matrix.
In particular, the identity matrix by itself ($1$'s down the main diagonal, $0$'s elsewhere) is not a subspace of the collection of $2\times 2$ matrices, for if the identity matrix $I$ is in the subspace, then $cI$ has to be in the subspace for all numbers $c$. The collection of all matrices which are $0$ off diagonal, and have all diagonal terms equal is a subspace of the space of all matrices. Maybe that will take care of part of your objection.
Let $V$ be any vector space, and take a collection $U$ of some of the elements of $V$. Then $U$ is called a subspace of $V$ if $U$ by itself is a vector space, meaning that the sum of any two elements of $U$ is in $U$, and any constant times an element of $U$ is in $U$.
You quoted something to the effect that a certain $D$ is a subspace of the space of upper triangular matrices. That's not true. The collection of all matrices of the shape you described, with everything off diagonal equal to $0$, is a subspace. So $D$ is supposed to be not a single matrix, it is a largish collection of matrices.
Now let's look at your particular problem. Let $V$ be the collection of all upper triangular matrices. Is this a vector space? Take any two upper triangular matrices $A$ and $B$. Is $A+B$ upper triangular? Yes. If $c$ is a constant, and $A$ is upper triangular, is $cA$ upper triangular? Yes. So $V$ is a vector space.
Let $D$ be the collection of all diagonal matrices? Is this a vector space? Yes, the sum of two diagonal matrices is diagonal, a constant times a diagonal matrix is a diagonal matrix. $D$ is a subspace of the upper triangular matrices, because any diagonal matrix is in particular upper triangular, it is a special upper triangular matrix.
Upper triangular matrices
$
\begin{bmatrix}
a&b\\
0&c
\end{bmatrix}
$
form a vector space with canonical basis:
$
e_1=\begin{bmatrix}
1&0\\
0&0
\end{bmatrix}
\quad
e_2=\begin{bmatrix}
0&1\\
0&0
\end{bmatrix}
\quad
e_3=\begin{bmatrix}
0&0\\
0&1
\end{bmatrix}
$
that is isomorphic to $\mathbb{R}^3$ by:
$
e_1\rightarrow\vec e_1=\begin{bmatrix}
1\\0\\0
\end{bmatrix}
\quad
e_2\rightarrow\vec e_2=\begin{bmatrix}
0\\1\\0
\end{bmatrix}
\quad
e_3\rightarrow \vec e_3=\begin{bmatrix}
0\\0\\1
\end{bmatrix}
$
Your linear transformation $T$ is defined as a matrix multiplication:
$
T(M)=\begin{bmatrix}
7&3\\
0&1
\end{bmatrix}
\begin{bmatrix}
a&b\\
0&c
\end{bmatrix}
=
\begin{bmatrix}
7a&7b+3c\\
0&c
\end{bmatrix}
$
so, in this canonical representation, it is given by the matrix $T_e$ such that:
$
T_e\vec M=\begin{bmatrix}
7&0&0\\
0&7&3\\
0&0&1
\end{bmatrix}
\begin{bmatrix}
a\\
b\\
c
\end{bmatrix}=
\begin{bmatrix}
7a\\
7b+3c\\
c
\end{bmatrix}
$
Now you want a representation of the same transformation in a new basis:
$
e'_1=
\begin{bmatrix}
1&0\\
0&0
\end{bmatrix}
\rightarrow\vec e'_1=\begin{bmatrix}
1\\0\\0
\end{bmatrix}
\quad
e'_2=\begin{bmatrix}
1&1\\
0&0
\end{bmatrix}\rightarrow\vec e'_2=\begin{bmatrix}
1\\1\\0
\end{bmatrix}
\quad
e'_3=\begin{bmatrix}
0&0\\
0&1
\end{bmatrix}\rightarrow \vec e'_3=\begin{bmatrix}
0\\0\\1
\end{bmatrix}
$
This transformation of basis is represented by the matrices
$
S=
\begin{bmatrix}
1&1&0\\
0&1&0\\
0&0&1
\end{bmatrix}
\qquad
S^{-1}=
\begin{bmatrix}
1&-1&0\\
0&1&0\\
0&0&1
\end{bmatrix}
$
So the matrix that represents the transformation $T$ in the new basis is:
$
T_{e'}=S^{-1}T_eS=
\begin{bmatrix}
7&0&-3\\
0&7&3\\
0&0&1
\end{bmatrix}
$
Best Answer
One source that I have has a definition (kind of hidden away in the questions): "An $m\times n$ matrix $A$ is called upper triangular if all entries lying below the diagonal entries are zero, that is, if $A_{ij}=0$ whenever $i>j$." (p.21 Friedberg et al, Linear Algebra 4th edition)
I have yet to find a source that explicitly contradicts this definition (so deliberately states that $m \times n$ matrices cannot be upper triangular), thereby limiting upper triangular matrices to square matrices only. But in all my other sources we have something similar to "...$A \in M_{n \times n}(K)$...upper triangular iff...". The other sources I could consult here was p.37 Cullen (Matrices and linear transformations) and p.149 Golan (The linear algebra a beginning graduate student ought to know).