Given a linear transformation $T$ on a finite-dimensional vector space $V$ over a field $K$, and giving $V$ the $K[x]$-module structure determined by $f(x)v = F(T)(v)$, Frederick Goodman shows on page 3 of this extract that: If $a(x)\in K[x]$ is monic with degree at least one and $V\cong K[x]/(a(x))$ as $K[x]$-modules, then $V$ has a basis with respect to which the matrix of $T$ is the companion matrix of $a(x)$. He begins this proof by stating that
[...] we may assume that $V$ is the $K[x]$-module $K[x]/(a(x))$, and that $T$ is the linear transformation $$ f(x) + (a(x)) \mapsto xf(x) + (a(x)). $$
It feels like circular reasoning to me that we give $V$ a $K[x]$-module structure, using the linear transformation $T$, and then redefine $V$ and $T$ in terms of this $K[x]$-module structure. I've seen this argument elsewhere so I think it's right, but I'm having a hard time justifying it to myself. How do we know that $T$ acts on $V$ in the same way after the redefining?
There is no redefinition. The $K[x]$ module structure on $K[x]$ is given by the usual multiplication. The $K[x]$ module structure on $K[x]/(a(x))$ is obtained by treating $K[x] / (a(x))$ as a quotient module so $f(x) (g(x) + (a(x))) = f(x)g(x) + (a(x))$. If $\varphi \colon V \rightarrow K[x]/(a(x))$ is an isomorphism of $K[x]$-modules, in particular it means that it is an isomorphism of vector spaces and that $T$ acts on $V$ just like $x$ acts on $K[x]/(a(x))$ because
$$ \varphi(T(v)) = \varphi(xv) = x\varphi(v). $$
In other words, if we denote by $L \colon K[x]/(a(x)) \rightarrow K[x]/(a(x))$ the linear map $L(f(x) + (a(x))) = xf(x) + (a(x))$ then we have $\varphi \circ T = L \circ \varphi$ or $T = \varphi^{-1} \circ L \circ \varphi$. Instead of working with $T$, the author suggests to work with $L$. If you prove that $L$ is represented with respect to some basis by a matrix then $T$ will be represented with respect to image of this basis under $\varphi$ by the same matrix.