Be $k$ a field, $k[x]$ the polynomial ring over $k$, $V$ a $k$-vector space, and $f \in \text{End}_k(V)$. We want to see how the $k$-vector space structure can be extended to a $k[x]$-module structure in a unique way such that $(\forall v \in V)(xv = f(v))$.
I thought I had the answer, but I'm running into some trouble at the end. This is my reasoning:
We already have a $k$-module $V$, i.e. a map $\phi : k \times V \rightarrow V$ that satisfies the module axioms. We define the following map: \begin{eqnarray} \psi : &k[x] \times V &\rightarrow V \\ &\left(\sum_{n=0}^∞ \lambda_nx^n, v\right) &\mapsto \phi(\lambda_0, v) + \sum_{n=1}^∞ \lambda_n f^n(v) \end{eqnarray} That way we'll have $\psi|_{k \times V} = \phi$ and $xv = \psi(x,v) = f(v)$. Checking the first two module axioms for $\psi$ is easy, but the third one ("$\psi(rs, v) = \psi(r, \psi(s,v))$") doesn't seem to work. On the one hand we have \begin{align*} \psi\left(\sum_{n=0}^∞ \lambda_nx^n \cdot \sum_{n=0}^∞ \mu_nx^n, v\right) &= \psi\left(\sum_{n=0}^∞\left( \sum_{i=0}^n\lambda_i\mu_{n-i}\right)x^n, v\right) \\ &= \phi(\lambda_0\mu_0, v) + \sum_{n=1}^∞ \left(\sum_{i=0}^n\lambda_i\mu_{n-i} \right)f^n(v) \\ &=\phi(\lambda_0, \phi(\mu_0, v)) + \sum_{n=1}^∞ \lambda_nf^n \circ \sum_{m=1}^∞ \mu_mf^m (v)\\ &=\phi(\lambda_0, \phi(\mu_0, v)) + \sum_{n=1}^∞ \sum_{m=1}^∞ \lambda_n\mu_mf^{n+m} (v). \end{align*}
On the other hand, we have \begin{align*} &\psi\left(\sum_{n=0}^∞\lambda_nx^n,\psi\left(\sum_{n=0}^∞\mu_nx^n, v\right)\right) \\ &= \psi\left(\sum_{n=0}^∞\lambda_nx^n, \phi(\mu_0, v)+ \sum_{n=1}^∞\mu_nf^n(v), \right) \\ &= \phi\left(\lambda_0, \phi(\mu_0, v) + \sum_{n=1}^∞\mu_nf^n(v)\right) + \sum_{n=1}^∞\lambda_n\cdot f^n\left( \phi(\mu_0, v)+ \sum_{n=1}^∞\mu_nf^n(v)\right) \\ &= \phi(\lambda_0, \phi(\mu_0, v)) + \phi\left(\lambda_0,\sum_{n=1}^∞\mu_nf^n(v)\right) + \sum_{n=1}^∞\lambda_n\cdot f^n\left( \phi(\mu_0, v)+ \sum_{n=1}^∞\mu_nf^n(v)\right) \end{align*}
We need these expression to be equal, but now we've introduced a term that seems irreconcilable with the above expression: $\phi\left(\lambda_0,\sum_{n=1}^∞\mu_nf^n(v)\right)$. Am I silently using a rule that is not justified here?
Let me simplify your notation. I will write $(\sum_{j=0}^n a_j x^j) \cdot v:= \sum_{j=0}^n a_j f^{j}(v)$. Note that $x^{j+k}(v) = f^{j+k}(v) = f^j (f^k(v)) = x^j \cdot (x^k \cdot v)$. Then we compute $$ \left(\left(\sum_{j=0}^n a_j x^j\right) \left( \sum_{k=0}^m b_k x^k \right) \right) \cdot v =\left(\sum_{j=0}^n \sum_{k=0}^m a_j b_k x^{j+k} \right)\cdot v = \sum_{j=0}^n \sum_{k=0}^m a_j b_k \left( x^{j+l} \cdot v \right) = \sum_{j=0}^n \sum_{k=0}^m a_j b_k x^j \cdot (x^k \cdot v) = \sum_{j=0}^n a_j x^j \cdot \left(\sum_{k=0}^m b_k x^k \cdot v\right) = \left( \sum_{j=0}^n a_j x^j \right) \cdot \left(\sum_{k=0}^m b_k x^k \cdot v\right) $$ Which is what we want to prove.
It is a fun exercise to think about when this $k[x]$-module will be free.