Let $G$ be a finite group. In M. Isaacs "Character theory of finite groups", chap. 2, we cover the fact that $\mathbb{C}[G] = \bigoplus_{i = 1}^{k} M_i(\mathbb{C}[G])$, where $M_i(\mathbb{C}[G])$ is the sum of all submodules of $\mathbb{C}[G]$ that are isomorphic to an irreducible $\mathbb{C}[G]$-module $M_i$. The reason for the finiteness is due to a theorem that gives us that the set $\mathcal{M}(\mathbb{C}[G])$ of isomorphism-classes of irreducible $\mathbb{C}[G]$-modules are finite.
Implicitly, we have here picked a representative set $\mathcal{M}(\mathbb{C}[G]) = \{M_1,\ldots,M_k\}$ of irreducible $\mathbb{C}[G]$-modules, one from each isomorphism-class.
Since $M_i$ is a $\mathbb{C}[G]$-module, it follows that $M_i$ is a $\mathbb{C}$-module (I believe), hence a $M_i$ is a vector space over $\mathbb{C}$. We can then pick a basis $\mathcal{B}_{M_i}$ for $M_i$, seen as a vector space over $\mathbb{C}$. As I understand it, we then get an induced representation $\mathfrak{X}_i$ on $M_i$, by how $G \subset \mathbb{C}[G]$ acts on the basis $B_{M_i}$ of $M_i$.
Furthermore, we have that $1 = \sum\limits_{i = 1}^{k} e_i$, where $e_i \in M_i(\mathbb{C}[G])$.
It is then claimed that $\mathfrak{X}_i(e_j) = 0$ if $i \neq j$. I am having trouble understanding what $\mathfrak{X}_i(e_j)$ means, since $\mathfrak{X}_i:G \to \text{GL}(M_i) \cong \text{Gl}(n,\mathbb{C})$ where $n$ is the dimension of $M_i$ as a $\mathbb{C}$-vector space. But $e_j$ is not neccessarily in $G$, right? So why do we assume that $\mathfrak{X}_i(e_j)$ is well-defined? I believe $e_j$ is called a central idempotent.
Edit: As @ahulpke pointed out in the comments, I believe we can extend $\mathfrak{X}_i$ to all of $\mathbb{C}[G]$ so that for $x = \sum\limits_{g \in G} a_{g}g \in \mathbb{C}[G]$ we define $\mathfrak{X}_i(x) = \mathfrak{X}_i\Big(\sum\limits_{g \in G} a_{g}g\Big) := \sum\limits_{g \in G} a_{g}\mathfrak{X}_i(g)$.
I am aware that Wedderburns theorem gives us that if $W$ is an irreducible $\mathbb{C}[G]$-module, then it is annihilated by $M_i(\mathbb{C}[G])$, unless $W \cong M_i$. But I am having trouble tying that to the representation $\mathfrak{X}_i$.
Edit 2: Is it as simple as the fact that we identify $e_j \cdot m_i := \mathfrak{X}_i(e_j) \cdot m_i$ for $m_i \in M_i(\mathbb{C}[G])$ so that since $e_j \cdot m_i = 0$ we have $\mathfrak{X}_i(e_j) \cdot m_i = 0$, hence $\mathfrak{X}_i(e_j) = 0$, since we are only interested in the action on $M_i$?
First (as discussed in the comments already), linear representations of the group extend naturally to the group algebra. Thus, it is sufficient for the theory to consider the representations of the algebra.
With this, here is a slightly different way of stating Wedderbuerns theorem over an algebraically closed field. It follows from what Isaacs proves, but he does not spell it out this way:
Let $A=\mathbb{F}G$ be a semisimple algebra over an algebraically closed field $\mathbb{F}$. Then $A$ is the direct sum of full matrix rings $\mathbb{F}^{n_i\times n_i}$, where the $n_i$ are the dimensions of the irreducible modules. That is, in a suitable basis (and finding this basis concretely is equivalent to finding the irreducible representations), the matrices for elements of $A$ in the regular (dimension $|G|$) representation are all the block diagonal matrices, with blocks of size $n_i\times n_i$, and all possible matrix entries in the blocks. In this form, the idempotent $e_i$ is the matrix that is zero, apart from an identity in the $i$-th block. The $i$-th irreducible representation consists simply of "cutting out" the $i$-th diagonal block.
The statement you wonder about follows immediately from this.