Components of Linear Operators

My book says that A linear operator $\mathcal{A}$ associates every vector $\vec{x}$ with another $\vec{y}$ such that you can say the $\mathcal{A}$ acts on $\vec{x}$ to give another vector $\vec{y}$ in the same basis ($\hat{e}_i$),

$$\tag{1}\label{1} \vec{y} = \mathcal{A}\vec{x}.$$

We can also say that the action of the linear operator $\mathcal{A}$ on the basis vectors $\hat{e}_i$ is independant and so we can write it as a linear combination of the basis vectors,

$$\tag{2}\label{2} \mathcal{A}\hat{e}_j=\sum_{i}^{N}\mathcal{A}_{ij}\hat{e}_i,$$

where $\mathcal{A}_{ij}$ is the $i^{\textrm{th}}$ component of the vector $\mathcal{A}\hat{e}_{j}$ in this basis. Together the elements $\mathcal{A}_{ij}$ are known as the components of the linear operator $\mathcal{A}$ in the $\hat{e}_i$ basis. The book then goes on to express $\vec{y} = \mathcal{A}\vec{x}$ in terms of these components,

$$\tag{3}\label{3} \vec{y}=\sum_{i}^{N} y_{i}\hat{e}_i = \mathcal{A} \left( \sum_{j}^{N} x_j\hat{e}_j\right) = \sum_{j}^{N}x_j \sum_{i}^{N} \mathcal{A}_{ij}\hat{e}_{i},$$

hence we can say,

$$ \tag{4}\label{4} y_i = \sum_{j}^{N}\mathcal{A}_{ij}x_j.$$

I am struggling to understand what $\eqref{2}$ through $\eqref{4}$ means. Part of my question is how to prove $\eqref{2}$ is true, if I can understand this the other steps follow. In addition, I am wondering if my intution about what this represents is right? I have been trying to think about this in terms $\mathcal{A}$ being a $NxN$ matrix with elements $\left\langle\hat{e}_i | \mathcal{A} |\hat{e}_j \right\rangle$, and multiplying it by a $Nx1$ column vector with elements $\left\langle\hat{e}_j|\vec{x}\right\rangle$. This will give me another set of vectors $\vec{y}$ with elements $\left\langle\hat{e}_i | \mathcal{A} |\hat{e}_j \right\rangle \left\langle\hat{e}_j|\vec{x}\right\rangle$. Is the above not just describing the action of multiplying an operator that can be represented as a matrix by a column vector so that each of the new elements $y_i$ in $\vec{y}$ is the inner product of the rows of the operator with the element $x_i$ in the vector $\vec{x}$? Is there a geometric meaning here?

I'm not sure I understand your confusion but I'll try to explain – let me know if this doesn't address your concerns.

$\mathcal{A}$ is a linear operator. That means it exists, independently of any basis. Indeed, you don't need a basis at all to define a linear operator.

(2) is defining what it means to represent $\mathcal{A}$ in terms of a matrix $A_{ij}$. It's a distinct concept – $A_{ij}$ is basis-dependent, whereas $\mathcal{A}$ is not. So you don't prove (2) – it's defining the matrix $A$.

This is sort of "deriving" matrix multiplication rules. So your intuition is right, but perhaps you're just looking at things the "wrong way round".