Find the cosets in $G/K$ I'm trying to find the cosets in
$G/K$ and write down the multiplication table of $G/K$ for
$G = ⟨a⟩ × ⟨b⟩$, where $o(a) = 8$ and $o(b) = 2$, and $K = \langle (a^2, b) \rangle$.
(In other words, $G=C_8×C_2$, the cartesian product of the cyclic groups of order 8 and 2, respectively)
I am familiar with the idea of cosets $Ha = \{ha \mid h \in H\}$ for right cosets generated by $a$ and $aH = \{ah \mid h \in H\}$ for left cosets generated by $a$.
Also to clarify, I am not referencing the $a$ used in the definition of group $G$ here, just giving an example of left and right coset definition.
I am not familiar with the process of finding cosets for quotient groups.
 A: EDIT.
Initially I interpreted the subgroup $K$ as being generated by two elements $a^2$ and $b$, answering accordingly. I have changed the answer to address the question where $K$ is generated by a single element $a^2b$.

First, the group $G$. If you think of $G$ as the product $C_8 \times C_2$, then the elements are of the form $(a,b)$, where
\begin{equation} 
\def\cor#1{\color{red}{#1}}
\def\cog#1{\color{darkgreen}{#1}}
\def\cob#1{\color{darkblue}{#1}}
\begin{aligned}
\cor{a^8} &\cor{{}= e} \qquad&& \text{($a$ has order $8$)}, \\
\cog{b^2} &\cog{{}= e} \qquad&& \text{($b$ has order $2$)}, \\
\cob{ab} &\cob{{}= ba} \qquad&& \text{($a$ and $b$ commute)}.
\end{aligned}
\label{rels}
\tag{ * }
\end{equation}
But you can also just use $a$ and $b$ as generators and multiply them in words like $a$, $ab$, $a^5$, $babbabab$, whatever.
These are equivalent by the correspondence (isomorphism of groups)
\begin{array}{c@{50em}cc}
G &\longleftrightarrow& C_8 \times C_2 \\
e && (e,e) \\
a && (a,e) \\
b && (e,b) \\
a^ib^j && (a^i,b^j) 
\end{array}
In $G$, the elements
\begin{array}{*8{l}}
e, &a, &a^2, &a^3, &a^4, &a^5, &a^6, &a^7, \\
b, &ab, &a^2b, &a^3b, &a^4b, &a^5b, &a^6b, &a^7b
\end{array}
are each distinct, and any word in $a$ and $b$ is equivalent to one of these by the relations (\ref{rels})
For example,
\begin{align}
\def\cor#1{\color{red}{#1}} 
\def\cog#1{\color{darkgreen}{#1}}
\def\cob#1{\color{darkblue}{#1}}
ba^{-6}b^{-1}ab^3
&= bea^{-6}b^{-1}ab^3
= b\cor{a^8}a^{-6}b^{-1}ab^3
= ba^2b^{-1}ab^2b \\
&= ba^2b^{-1}a\cog{e}b 
= ba^2b^{-1}ab
= \cob{a^2b}b^{-1}ab 
= a^2eab
= a^2ab
= a^3b
\end{align}
which is one of those $8 \cdot 2 = 16$ elements of $G$
Notice how the exponents in $a^ib^j$ are just the totals for each of the exponents in the original words:
$$
i = (-6)+1 = -5 \equiv 3 \pmod{8} 
\qquad\text{and}\qquad 
j = 1+(-1)+3 = 3 \equiv 1 \pmod{2}
$$

Now the subgroup $K = \langle a^2b \rangle < G$.
Let's work out the left cosets of $K$. (But since this group is abelian, these are also the right cosets.)
Since $K$ is generated by $a^2b$, elements of the subgroup consist of powers of $(a^2b)^i = a^{2i}b^{i}$, where the exponent of $a$ is considered mod $8$ and the exponent of $b$ is considered mod $2$, i.e.
\begin{array}{*8{l}}
\def\coy#1{\color{lightgray}{#1}}
e, &\coy{a,} &\coy{a^2,} &\coy{a^3,} 
&a^4, &\coy{a^5,} &\coy{a^6,} &\coy{a^7,} \\
\coy{b,} &\coy{ab,} &a^2b, &\coy{a^3b,} 
&\coy{a^4b,} &\coy{a^5b,} &a^6b, &\coy{a^7b}
\end{array}
The subgroup $K$ is the identity coset, but any of these elements can be a representative of the coset (it goes by many names):
$$
K = eK = a^2bK = a^4K = a^6bK.
$$
Now, consider the coset $aK$, consisting of elements $a^{2i+1}b^i$:
\begin{array}{*8{l}}
\def\coy#1{\color{lightgray}{#1}}
\coy{e,} &a, &\coy{a^2,} &\coy{a^3,} 
&\coy{a^4,} &a^5, &\coy{a^6,} &\coy{a^7,} \\
\coy{b,} &\coy{ab,} &\coy{a^2b,} &a^3b, 
&\coy{a^4b,} &\coy{a^5b,} &\coy{a^6b,}, &a^7b
\end{array}
This coset also has $4$ names, using any of the elements as a representative:
$$
aK = a^3bK = a^5K = a^7bK.
$$
There are $2$ more cosets:
$$
a^2K = a^4bK = a^6K = bK, 
$$
and
$$
a^3K = a^5bK = a^7K = abK. 
$$
That's it! Since $\lvert G \rvert = 16$ and $\lvert K \rvert = 4$, we only expect $16/4 = 4$ cosets. In other words, the subgroup has index $4$, denoted $[G: K] = 4$.
This means the quotient group $G/K$ has order $4$. Explicitly, the elements of $G/K$ are $\bigl\{ K, aK, a^2K, a^3K \bigr\}$, and the multiplication table looks like:
\begin{array}{r|r*3{@{2em}r}}
   &  K & aK & a^2K & a^3K \\
\hline
   K &    K &   aK & a^2K & a^3K \\
  aK &   aK & a^2K & a^3K &    K \\
a^2K & a^2K & a^3K &    K &   aK \\
a^3K & a^3K &    K &   aK & a^2K
\end{array}
You don't have to take my word for it. For example, since a typical element in $K$ looks like $a^{2i} \, b^i$, where $i \in \mathbb{Z}$, a typical element in $a^mK$ looks like $a^{2i+m} \, b^i$, so we can compute, e.g., $a^mK \, a^nK$ by multiplying generic elements in the cosets:
$$
\bigl( a^{2i+m} b^i \bigr) \, 
\bigl( a^{2j+n} b^j \bigr) 
= a^{2(i+j)+m+n} \, b^{i+j} 
= a^{m+n} \, a^{2(i+j)} \, b^{i+j}, 
$$
which is in $a^{m+n}K$, where the power is considered mod $4$ (since $a^4K = K$).
If this quotient group looks like the cyclic group of order $4$, that's because it is isomorphic! To make this explicit, say $C_4 = \langle c \rangle = \{e, c, c^2, c^3\}$, where $c^4=e$. Then the correspondence is
\begin{array}{c@{50em}cc}
G/K &\longleftrightarrow& C_4 \\
a^mK && c^m 
\end{array}

Putting all this together, the quotient group is
$$
G/K \cong C_4.
$$
