Consider the linear transformation $T$ that takes a vector in $(x,y)\in\mathbb{R}^2$ and maps it to the vector $(x+2y,3x+4y)\in\mathbb{R}^2$. That is,
$$
\begin{bmatrix}
x\\
y
\end{bmatrix}
\overset{T}{\longmapsto}
\begin{bmatrix}
x+2y\\
3x+4y
\end{bmatrix}.
\tag{1}
$$
We want to express this transformation in the form of $A\mathbf{x}$ where $A$ is a matrix and $\mathbf{x}=(x,y)$ is the input vector. Hopefully you can see here that the matrix of the linear transformation described in $(1)$ is
$$
A=
\begin{bmatrix}
1 & 2\\
3 & 4
\end{bmatrix}
$$
since
$$
A\mathbf{x}=\underbrace{\begin{bmatrix}
1 & 2\\
3 & 4
\end{bmatrix}}_{\text{matrix of }T}\ \
\underbrace{\begin{bmatrix}x\\
y\end{bmatrix}}_\text{input vector}
=
\underbrace{\begin{bmatrix}
x+2y\\
3x+4y
\end{bmatrix}
}_\text{output vector}
$$
Note that $A$ is not the linear transformation per se; rather $A$ is a computationally efficient way to write down the action taking place in $(1)$ in the form of the matrix-vector multiplication $A\mathbf{x}$.
Implicit in this conversation is that we are writing down the matrix of the linear transformation in $(1)$ with respect to some fixed basis. If changing the basis is your emphasis rather than the general idea of the connection between a linear transformation and its associated matrix, you might study this.
