In the book The Symmetric Group by Sagan the author defines a $G$-module to be a vector space $V$ that has a multiplication $g\mathbf{v}$ of elements of $g\in G$ with elements of $\mathbf{v}\in V$ such that
- $g\mathbf{v}\in V$
- $g(c\mathbf{v}+d\mathbf{w})=c(g\mathbf{v})+d(g\mathbf{w})$,
- $(gh)\mathbf{v}=g(h\mathbf{v})$
- $\epsilon\mathbf{v}=\mathbf{v}$ for all $g,h\in G; \mathbf{v},\mathbf{w}\in V$; and $c,d$ in the field.
He later writes that if we have a matrix representation $X$ of degree $d$ and let $V$ be a vector space of column vectors of length $d$, then we can define a $G$-module by \begin{equation}\tag{1} g\mathbf{v}:=X(g)\mathbf{v},\qquad g\in G,\mathbf{v}\in V. \end{equation} However, I try to see how this can fulfill condition 3 above (associativity), but it does not seem to add up. Because eq. $(1)$ is the same as saying that for each $v_i$ in the basis, we have $$\tag{2} gv_i=\sum_j X_{ij}(g)v_j $$ But since X is a representation we have $X(gh)=X(g)X(h)$ and thus: $$\tag{4} (g h) v_i=\sum_j X_{i j}(g h) v_j=\sum_{j, k} X_{i k}(g) X_{k j}(h) v_j $$ $$\tag{5} g\left(h v_i\right)=g \sum_k X_{i k}(h) v_k=\sum_k X_{i k}(h) g v_k=\sum_{j, k} X_{i k}(h) X_{k j}(g)v_j $$ So associativity does not hold. To me it seems like eq. $(1)$ should really be defined $$ g\mathbf{v}:=\mathbf{v}^TX(g) $$ but I could be wrong.
Looking into the book, we have that $V = \mathbb{C}^d$ is the space of all column vectors with entries in $\mathbb{C}$. $X : G \rightarrow Gl_d(\mathbb{C})$ is a homomorphism (viewing the one on the right as a group under composition). In particular, this means that $X(gh) = X(g) X(h)$, by definition of a homomorphism. Notice that this induces a map $X_{ij} : G \rightarrow \mathbb{C}$ which is the $ij$ component of the matrix $X(g)$. The homomorphism property tells us $$X_{ij} (gh) = \sum_{k=1}^d X_{ik}(g) X_{kj}(h).$$ Notice also we can write the group action as $$(X(g)v)_i = \sum_{j=1}^d X_{ij}(g) v_j.$$ I think here lies the issues with your calculations - you treat $v_j \in V$ instead of a complex number. Here we're thinking of it as a number.
Putting this together, $$(X(gh)v)_i = \sum_{j=1}^d X_{ij}(gh) v_j = \sum_{j,k=1}^d X_{ik}(g) X_{kj}(h) v_j = \sum_{k=1}^d X_{ik}(g) (X(h)v)_k = (X(g) (X(h)v))_i. $$ We get the same vector on both ends.
If V is a general $\mathbb{C}$ vector space of dimension $d$, then we have an isomorphism between $V$ and $\mathbb{C}^d$ (what might that be?) and we're using this identification when talking about things via matrices (most likely one doesn't have to use this identification but for the sake of clarity we might as well).
Edit: To emphasize my point, write out your vector $v = \sum_{i=1}^d v_i e_i$, where $e_i$ is a basis. Now
$$X(g)v = X(g)\left( \sum_{j=1}^d v_i e_i \right) = \sum_{i=1}^d v_i X(g)(e_i).$$
Recall that the matrix $X(g)$ acts on the vector $e_i$ by $X(g)(e_j) = \sum_{i=1}^d X_{ji}(g)e_i.$ So we get
$$X(g)v = X(g)\left( \sum_{i=1}^d v_i e_i \right) = \sum_{i=1}^d v_i \sum_{k=1}^d X_{ki}(g) e_k = \sum_{k=1}^d \left(\sum_{i=1}^d X_{ki}(g) v_i \right) e_k.$$
If we project onto the subspace spanned by $e_k$, we get
$$ (X(g)v)_k = \sum_{i=1}^d X_{ki}(g) v_i,$$
which matches my formula from earlier. Now try the same calculations with this.