Yes, take cosets $A=aK$, $B=bK$, then the first definition
$$
A\cdot B := (ab)K
$$
is a coset again, by definition, but we have to check that the choice of representatives $a\in A$ and $b\in B$ is irrelevant.
For the second definition,
$$
A\cdot B := AB = \{\, gh : g\in A, h\in B\,\},
$$
you don't have to check any independence of choice of representatives, because the definition doesn't invole representatives. Instead, you have to check that what you get is a coset again, and as it turns out, you get $AB=(ab)K$, so both definitions agree.
Be aware that you need normality of $K$ in both cases to get something that is well-defined and a coset again.
I'll do the first one for you.
$H=<(123)>=\{(1),(123),(132)\}$. A left coset of $H$ is a set of the form $aH$, where $a\in A_{4}$. One thing you probably learned is that two left cosets are equal to one another or disjoint. Note that
$$
\begin{aligned}
&(12)(34)H=\{(12)(34), (243),(143)\}\\
&(13)(24)H=\{(13)(24), (142), (234)\}\\
&(14)(23)H=\{(14)(23), (134), (124)\}.
\end{aligned}
$$
Thus,
$$
A_{4}=H\sqcup(12)(34)H\sqcup(13)(24)H\sqcup(14)(23)H,
$$
where $\sqcup$ denotes a disjoint union.
Similarly, for right cosets, we have
\begin{aligned}
&H(12)(34)=\{(12)(34), (134),(234)\}\\
&H(13)(24)=\{(13)(24), (243), (124)\}\\
&H(14)(23)=\{(14)(23), (142), (143)\}.
\end{aligned}
Thus,
$$
A_{4}=H\sqcup H(12)(34)\sqcup H(13)(24)\sqcup H(14)(23).
$$
Now, note that $(12)(34)H\not=H(12)(34)$. Thus, $H$ is not normal in $A_{4}$.
Recall that when computing cosets, the representative is not unique. That is, I used $(12)(34),(13)(24), (14)(23)$ and $(1)$ as the coset representatives. The idea is you start with $H$. Then, to get a different coset, choose $a\in A_{4}\setminus H$ to get $aH$. Then, to get another coset, choose $b\in A_{4}\setminus(H\cup aH)$, and so on, until you have all of the elements in $A_{4}$ contained in one of the cosets.
Best Answer
You are correct arguing that cosets are equicardinal (the same size), since for fixed $x$ the map $n \to xn$ is a bijection.
Your mistake (pointed out in comments and the other answer) is saying that makes them isomorphic. They can't be, since only $N$ itself is a group. When you are studying group theory, "isomorphic" means "isomorphic as groups".
In general, isomorphisms preserve the structure of the objects you are studying. When you are studying sets, an isomorphism is just a bijection, since sets have no other structure. So the map $n \to xn$ is indeed a set isomorphism.