Let $G=C_{p^{k_1}}\times C_{p^{k_2}}\times ... \times C_{p^{k_n}}$ an abelian $p$ group, while $k_1,...,k_n\in\mathbb{N}$ and $k_1\geq k_2 \geq ...\geq k_n$. A group $U\cong C_{p^{l_1}}\times C_{p^{l_2}}\times ... \times C_{p^{l_n}}$ is isomorphic to a subgroup of $G$ if and only if $l_1,...,l_n\in\mathbb{N}_0$ and $l_1\leq k_1,..., l_n\leq k_n$.
Now, I have a group $G=C_8\times C_2$ and I have to show that $G$ has an $C_4$ isomorphic subgroup $U$ so that $G/U$ is isomorphic to $C_4$. How can I prove it?
Your statement about subgroups proves the first statement which you seek to prove. To make this obvious consider $C_8=C_{2^3}$ which is of the form $C_{p^k}$. Next consider $C_4=C_{2^2}$ which is also of the form $C_{p^l}$. Now consider $C_1$ which trivially is of the form $C_{p^0}$. Finally we consider that $C_4 \cong C_4 \times C_1$ provided by the map $C_4$ to $C_4$; and $C_4$ to $C_1$. Obviously $3 \gt 2$ and $1 \gt 0$, thus by your prior statement you must conclude $C_4 \lt G$. (really $C_4 \times C_1 \lt G$ but we know that is isomorphic to just $C_4$)
Note: we have proven this with out actually showing a way to find explicit elements of the subgroup $U$ from $G$, this requires some way to determine the subgroups ie. a homomorphism.
There are two ways of going about the second question, one to construct the groups in question, or another way which is to consider more general theorems about abstract algebra and really that follow from category theory. I shall not do the first one. Instead If you consider the isomorphism theorems for groups, they say that for any group $G$ if there is a homomorphism $\phi$ then the kernal $kern(\phi)$ is a normal subgroup and $ G/kern(\phi)$ is isomorphic to the image of $\phi$. So $ 8 \Bbb Z \to 4\Bbb Z$ by $a \mapsto a\;mod\;4$. Clearly this is a homomorphism so by the isomorphism theorems $G/U \cong C_4$.