Clarification of notation on vector space $V$ as a $F[x]$-module. I need some clarification on notation. In Advanced Linear Algebra they write

If $\tau\in\mathcal{L}(V)$ we will think of $V$ not only as a vector space over a field $F$ but also as a module over $F[x]$, with scalar multiplication defined by
  $$p(x)v=p(\tau)(v).$$

Clearly $p(x)$ is meant as a polynomial in $F[x]$. Fairly evident as such but I am curious about the $p(\tau)$ as for a polynomial we have
$$p(x)=\sum_n a_n x^n$$
where I leave out the index limits for simplicity and putting in $\tau$ seems a bit odd, I presume what they mean with $\tau^n$ then is repeated composition, that is $\tau^2=\tau\circ\tau$, as any form of multiplication would be meaningless there I feel. As afterward we have a vector and otherwise it doesn't make much sense, so I interpret it as
$$p(\tau)(v) = \sum_n a_n \tau^n(v)$$
where $\tau^n$ is again the repeated composition, is this correct interpretation? 
 A: Your interpretation is correct, but it requires some work to achieve a formal justification. It may seem proving the obvious, but it isn't really.
The set $\operatorname{End}_F(V)$ of $F$-linear endomorphisms of $V$ is a ring under endomorphism addition and composition. If we denote by $\mathbf{1}$ the identity endomorphism, for a scalar $\alpha$ and $\tau\in\operatorname{End}_F(V)$ we have $(\alpha\mathbf{1})\tau=\tau(\alpha\mathbf{1})$ (and we can also write $(\alpha\mathbf{1})=\alpha\tau$ with the usual definition of multiplication of an endomorphism by a scalar). This is usually referred to as $\operatorname{End}_F(V)$ being an $F$-algebra: the field $F$ is naturally embedded in the ring $\operatorname{End}_F(V)$.
Now an important result in abstract algebra is that whenever $A$ is an $F$-algebra, an $F$-algebra homomorphism $F[x]\to A$ is completely determined by assigning an image to $x$.

If $a\in A$, then there exists a unique $F$-algebra homomorphism $\varepsilon_a\colon F[x]\to A$ such that $\varepsilon_a(x)=a$.

What is this homomorphism? It obviously should be
$$
\varepsilon_a(\alpha_0+\alpha_1x+\dots+\alpha_nx^n)=
\alpha_0\mathbf{1}+\alpha_1a+\dots+\alpha_na^n
$$
(where $\mathbf{1}$ denotes the identity in $A$). The boring part of the proof is showing this is indeed an $F$-algebra homomorphism (and I'll skip it).
In our case $A=\operatorname{End}_F(V)$ and $a=\tau$. Since multiplication is map composition, you see your guess was right.
There is no problem in showing now that $V$ becomes an $F[x]$-module: it is, essentially by definition, a left module over $\operatorname{End}_F(V)$ and so we can use the $F$-algebra homomorphism $\varepsilon_\tau$ for inducing the $F[x]$-module structure: for $p\in F[x]$ and $v\in V$,
$$
pv=\varepsilon_\tau(p)v
$$
and in the right-hand side we have precisely the action of $\varepsilon_\tau$, as endomorphism of $V$, on $v$.
