$\newcommand{\R}{\mathbb{R}}$ tl;dr: In what way is $w = T(v)$ "the same" as $[w]_C = {}_C [T]_B [v]_B$? Here $V,W$ are vector spaces, $B,C$ are respective bases for them, and $v \in V$, $w \in W$, $T \in \operatorname{Hom}(V,W)$.
Background
Let $V,W$ be finite-dimensional vector spaces (over $\R$, say) of dimension $n,m$, respectively. By choosing bases $B,C$ of these spaces, we obtain isomorphisms $f:V \rightarrow \R^n$ and $g:W \rightarrow \R^m$, but we usually denote the images of these maps by $[v]_B$ and $[w]_C$ instead of $f(v)$ and $g(w)$.
Moreover, if we have a map $T \in \operatorname{Hom}(V,W)$, we can represent it as a matrix ${}_C [T]_B \in M_{m \times n}(\R)$ such that if $w = T(v)$, then:
$$ [w]_C = {}_C [T]_B [v]_B $$
In particular, the map $\operatorname{Hom}(V,W) \rightarrow M_{m \times n}(\R) : T \mapsto {}_C [T]_B$ is an isomorphism.
In short, choosing bases $B$ and $C$ produces isomorphisms $V \rightarrow \R^n$, $W \rightarrow \R^m$, and $\operatorname{Hom}(V,W) \rightarrow M_{m \times n}(\R)$. However, it does more than that: it preserves a certain operation between these spaces. On one hand we have application of a function to a vector space, and on the other we have matrix multiplication, and these operations work the same way when we identify the elements correctly. It's like we have an operator $\mathcal{M}$ which works like this:
$$ \mathcal{M}\{w\} = \mathcal{M}\{T(v)\} = \mathcal{M}\{T\} \mathcal{M}\{v\} $$
This is very reminiscent of a group homomorphism, and yet this is not what it is since the elements $\mathcal{M}$ applies to belong to different spaces. However, it is suggestive of a larger algebraic structure in which all this is happening, and with respect to which "linear maps" and "matrix multiplication" are the same (or, if you will, isomorphic).
My question is: what algebraic structure (if any) encapsulates this idea?
$\newcommand{\R}{\mathbb{R}}$ Here is a structure that I created, which feels a bit contrived.
A linear mapspace over a field $\mathbb{F}$ is a triple $(V,M,W)$ of vector spaces over $\mathbb{F}$, together with a bilinear operation $\cdot : M \times V \rightarrow W$.
Given two linear mapspaces $(V_1, M_1, W_1)$ and $(V_2, M_2, W_2)$, a linear mapspace homomorphism is a triple $(f_V, f_M, f_W)$ of linear functions $f_V : V_1 \rightarrow V_2$, $f_M : M_1 \rightarrow M_2$, and $f_W : W_1 \rightarrow W_2$, such that:
$$ f_W(m \cdot v) = f_M(m) \cdot f_V(v) \qquad \qquad \forall v \in V_1, \forall m \in M_1 $$
EDIT: Concretely, $(\R^n, M_{m \times n}(\R), \R^m)$ is a linear mapspace. If $V,W$ are vector spaces, so is $(V, \operatorname{Hom}(V,W), W)$. Moreover, if $V,W$ have dimension $n,m$, and we pick bases $B,C$ for them, then the triple:
$$ \left(v \mapsto [v]_B, T \mapsto {}_C[T]_B, w \mapsto [w]_C \right) $$
Is a linear mapspace homomorphism from $(V, \operatorname{Hom}(V,W), W)$ to $(\R^n, M_{m \times n}(\R), \R^m)$.