I'm having trouble linking the notation of something like $\mathbb{Z}/n$ and $R/H$ where $n \in \mathbb{Z}$, $R$ is a ring, and $H$ is the kernel of some homomorphism from that ring to another.
In $\mathbb{Z}/n$, we effectively set $n$ to zero, and express every element of $\mathbb{Z}$ as $m < n$ plus some multiple of $n$. (So, consider $\mathbb{Z}/3 = \mathbb{Z_3} = \{0, 1, 2\}$—since 3 is now the additive identity, every element is either 0, 1, or 2 added to some multiple of 3 [which is zero]).
On the other hand, in $R/H$, we effectively set elements of H to zero—but elements of $R/H$ are the additive cosets of $H$. So every element in $R$ is $\{r + h_1, r + h_2, ... , r + h_n\}$. Is the notation convention here just inconsistent? Or have I not read far enough?
I suppose every element in $R$ is some multiple of an element in the kernel added to something less than said element, but why isn't it written that way? Just for ease-of-access? (At least, my book says: Elements of $R/H$ are the additive cosets of $H$.)
Well, the idea is to assume that for a ring $R$ and this special subset $H$, we want $H$ to be an ideal of $R$, that is, $$ \forall x,y \in H, \forall r \in R, \quad x+y, rx \in H. $$ Then, for each element $r \in R$, we group together all the elements $\{ r+h \, | \, h \in H \}$. Note that $H$ doesn't have to be finite as in the notation you suggested by writing $\{ r + h_1, \cdots, r+h_n \}$.
Usually, one denotes $$ \overline r \overset{def}= \{ r+h \, | \, h \in H \}. $$ Note that with this notation, $\overline{r} = \overline{r+h}$ for any $h \in H$.
In the example of $\mathbb Z / 3 \mathbb Z$, this means $$ \overline 0 = \overline 3 = \overline 6 = \cdots = \overline{3n} $$ for any $n \in \mathbb Z$. But OF COURSE, $0 \neq 3$. Contrary to what you indicated in the question, $\mathbb Z / 3 \mathbb Z \neq \{0,1,2\}$, but rather $\{\overline 0, \overline 1, \overline 2\}$. The elements of $\mathbb Z / 3 \mathbb Z$ are really the cosets, but since looking at these sets with infinitely many elements is kind of heavy, we use representatives for the cosets ; that is, $\overline r = \{ r +h \, | \, h \in h \}$ admits the representative $r$, and $\{ \cdots,-6,-3,0,3,6,9,\cdots \}$ admits the representative $0$, $3$, and so on. Any element of the coset can be chosen as a representative, since $\overline r = \overline{r+h}$ for any $ h \in H$.
So in informal terms "we do set $n$ to zero", but in formal terms, "we put $0$ and $n$ in the same coset". The point of this construction is that since $H$ is an ideal, addition and multiplication are well-defined on the representatives ; we can define $$ \overline r + \overline s \overset{def}= \overline{r+s}, $$ because if $r+h$ and $s+h'$ are other representatives for the sets $\overline r$ and $\overline s$, then $$ (r+h) + (s+h') = (r+s) + (h+h'), $$ and so the function defined above does not depend on the choice of representative. Similarly for multiplication : $$ \overline{r} \, \overline{s} = \overline{rs} $$ is well-defined since $$ (r+h)(s+h') = rs + \underset{\in H}{\underbrace{(hs + rh' + hh')}}. $$ This explains why we can write $\overline 2 \cdot \overline 2 = \overline 4 = \overline 1$.
Hope that helps,