Let Z define the integers and $Z_a$ define the integer group modulo a. I want to determine what A is.
Given $Z_4\cong Z_8\oplus Z_2/A$, where $A\subset Z_8\oplus Z_2$, am I able to just say that given $H=\{z_0, z_4\} \subset Z_8\oplus Z_2$ where $z_0=id$ and $(z_4)^2=id$, $A\cong H \oplus Z_2$? Im am doing this just by observation so I am a bit sceptical. Are there cases where this isnt true?
i.e. If $Z_4\cong Z_8\oplus Z_2/A$ but A is defined differently to how I did it before?
Well, the notation is a bit dangerous, since you seem to be conflating $=$ with $\cong$. Other than that, we know that any subgroup of $Z_8\oplus Z_2$ must be of the form $B\oplus C$, where $B\le Z_8$ and $C\le Z_2$. So for $C$ our only choices are the trivial group and the whole group. Also note that $|Z_4|=4$,$|Z_8\oplus Z_2|=16$, hence $|A|=|B||C|=4$. So, if $|C|=1$, we have $|B|=4$, and the only subgroup of order $4$ of $Z_8$ is the one generated by the generator squared....then we would have $Z_8\oplus Z_2/A\cong Z_2 \oplus Z_2$ which is false, hence $|C|=2=Z_2$. Then $|B|=2$, which is the only subgroup that you identified.
In other words, your $A$ is correct and unique