It is easy to see that the set of real sequences $\mathbb{R}^{\mathbb{N}}$ is a ring. It suffices to define, for all $r,s\in\mathbb{R}^{\mathbb{N}}$, the operations $r\oplus s =(r_n+s_n)_{n\in\mathbb{N}}$ and $r\odot s=(r_n\cdot s_n)_{n\in\mathbb{N}}$.
Let $\mathcal{U}$ be a nonprincipal ultrafilter on $\mathbb{N}$. For all $r\in\mathbb{R}^{\mathbb{N}}$, we define the set $r^{(0)}=\{n\in\mathbb{N} \mid r_n=0\}$.
My question: Is the set of the $\textit{almost null sequences}$ $$ \mathbb{I} = \{r\in\mathbb{R}^{\mathbb{N}}\mid r^{(0)}\!\in\mathcal{U}\} $$ a two-sided ideal of $\mathbb{R}^{\mathbb{N}}$? I think yes, because if $s\in\mathbb{I}$ and $r\in\mathbb{R}^{\mathbb{N}}$, then $(s\odot r)^{(0)}\in\mathcal{U}$ (i.e. the product of any sequence and an almost null sequence is almost null).
If yes, is the set of the hyperreal numbers the quotient ring $\mathbb{R}^{\mathbb{N}}\diagup \mathbb{I}$? In this case two sequences should belong to the same class if their difference is almost null (namely, they match on an index set which belongs to the ultrafilter $\mathcal{U}$)
Yes, $\mathbb{I}$ is an ideal. One needs to check closure under addition, but this is straightforward: The intersection of two sets in the ultrafilter is in the ultrafilter.
The quotient construction you described is correct, it is the usual construction of an ultrapower. It is one of the ways to construct a non-standard model of analysis. There are others: a non-standard model of analysis is not necessarily an ultrapower. Also, the index set need not be the natural numbers. It is not really accurate to speak of the hyperreals, since there are non-standard models of analysis of arbitrarily large cardinality.
Note that the hyperreals are much more than a field. In particular, the ultrapower construction gives, for each function $f:\mathbb{R}^n\to\mathbb{R}$, and for every relation $A\subset \mathbb{R^n}$, a natural extension to the ultrapower that preserves first-order properties.