Let $R$ be a ring and $M$ an abelian group. To give $M$ the structure of a left $R$-module is precisely to give a ring homomorphism $R \to \text{End}(M)$, where $\text{End}(M)$ denotes the ring of endomorphisms $M \to M$.
If $R = k$ is a field, giving a ring homomorphism $k \to \text{End}(M)$ equips $M$ with a notion of scalar multiplication by elements of $k$, so in fact a $k$-module is the same thing as a $k$-vector space.
If $R = k[x]$, then a ring homomorphism $\phi : k[x] \to \text{End}(M)$ gives by precomposition a ring homomorphism $k \to k[x] \to \text{End}(M)$, so $M$ is in particular a $k$-vector space. In addition, we get an additive map $\phi(x) \in \text{End}(M)$, that is, one satisfying $\phi(x)(m + n) = \phi(x)m + \phi(x)n.$
Since $x$ commutes with all scalars in $k$, we have in fact that $\phi(x)$ is an endomorphism of $M$ as a $k$-vector space; moreover, the image of any element of $k[x]$ is determined by the structure of $M$ as a $k$-vector space and by $\phi(x)$. So we conclude that
A $k[x]$-module is precisely a $k$-vector space together with a linear operator.
So I do not really know how to answer your questions as they are phrased. Rotman is just unpacking the definition of a $k[x]$-module. To me the question you are asking is roughly analogous to the following question:
My calculus teacher gave me the following construction of the derivative of a polynomial: to a polynomial $\sum a_i x^i$ you associate the polynomial $\sum i a_i x^{i-1}$. Is this construction important enough to have a name and, if so, what is it called and how does it fit into the bigger picture? What is the motivation for this construction?
Your calculus teacher is just unpacking the definition of differentiation.