The following is taken from "Module Theory an approach to linear algebra" by T.S Blyth
$\color{Green}{Background:}$
$\textbf{(1) Example}$
Let $R$ be a unitary ring and let $R^N$ denote the set of all mappings $f:N\to R$ (i.e., the set of all sequences of elements of $R$). Endow $R^N$ with the obvious addition, namely $f,g\in R^N$ define $f+g:N\to R$ by the prescription
$$(f+g)(n)=f(n)+g(n).$$
Clearl, $R^N$ forms an abelian group under this law. Now define an external law $R\times R^N\to R^N$ by $(r,f)\mapsto rf$ where $rf:N\to R$ is given by the prescription
$$(rf)(n)=rf(n).$$
This then makes $R^N$ into an $R-module.$
$\textbf{(2) Example}$
Let $R$ be a commutative unitary ring and consider the $R-$module $R^N$ of $\textbf{(1) Example}.$ Given $f,g\in R^N,$ define the product map $fg:N\to R$ by the prescription
$$(fg)(n)=\sum_{i=0}^{n}f(i)g(n-1).$$
Then it is readily verified that the law of composition described by $(f,g)\mapsto fg$ makes $R^N$ into an $R-$algebra. This $R-$algebra is called the $\textit{algebra of formal power series with coefficients in R.}$ The reason for this traditional terminology is as follows. Let $t\in R^N$ be given by
$$t(x)=\begin{cases} 1 \quad &\text{if} \, n=1; \\ 0 \quad &\text{otherwise.}\\ \end{cases}$$
Then for every positive integer $m$ the $m-$fold composite map $t^m=t\circ t\circ\ldots\circ t$ is given by
$$t^m(x)=\begin{cases} 1 \quad &\text{if} \, n=m; \\ 0 \quad &\text{otherwise.}\\ \end{cases}$$
Consider now (without worrying how to imagine the sum of an infinite number of elements of $R^N$ or even questioning the lack of any notion of convergence) the 'formal power series' associated with $f\in R^N$ given by
$$\theta=f(0)t^0+f(1)t^1+f(2)t^2+\ldots +f(m)t^m\ldots$$
where $t^0=\mathrm{id}_R,$ the identity map on $R.$ Since $\theta(n)=f(n)$ for every $n\in N$ it is often said that '$f$ can be represented symbolically by the above formal power series'.
$\color{Red}{Questions:}$
I am a bit confused about the notation: $\theta=f(0)t^0+f(1)t^1+f(2)t^2+\ldots +f(m)t^m\ldots$ for $\theta(m),$ specifically, how $t^m(x)$ relates to $t^m$ in $\theta.$ So say we have $m=5$, then $t^5(x)=\begin{cases} 1 \quad &\text{if} \, n=m; \\ 0 \quad &\text{otherwise.}\\ \end{cases}.$ We would have $\theta=f(0)t^0+f(1)t^1+f(2)t^2+f(3)t^3+f(4)t^4+f(5)t^5.$ But $\theta(n)=f(n)$ only when $n=5.$ This would make $\theta(5)=f(0)t^0+f(1)t^1+f(2)t^2+f(3)t^3+f(4)t^4+f(5)t^5=f(5)t^5.$ To get the entire formal power series $f(0)t^0+f(1)t^1+f(2)t^2+\ldots +f(m)t^m\ldots,$ We would need $f(0)t^0+f(1)t^1+f(2)t^2+\ldots +f(m)t^m\ldots=\theta(0)t^0+\theta(1)t^1+\theta(2)t^2+\ldots +\theta(m)t^m\ldots,$ because the $t^m-$s in $f(m)t^m$ stands for $t^m(n).$ Am I correct in my understanding of the notation?