I know that
$$ \mathbb{Z}_2 = \{0, 1\} \\ \mathbb{Z}_4 = \{0, 1, 2, 3\} $$
Then
$$ \mathbb{Z}_2 \oplus \mathbb{Z}_4 = \{ (0,0), (0,1), (0,2), (0,3), (1,1), (1,2), (1,3) \} $$
We want to find all subgroups of this with order 4. By Lagrange's theorem, the order of every subgroup must divide the order of $\mathbb{Z}_2 \oplus \mathbb{Z}_4$, namely 7. Thus each subgroup must have order of 1. Therefore, no subgroup exists with order 4.
Is this correct?
The order of $G=\mathbb{Z}_2\oplus\mathbb{Z}_4$ is not $7$, it is $8$. Remember that for the Cartesian product we have
$$|X\times Y|=|X|\cdot |Y|$$
And the direct sum "$\oplus$" is just the Cartesian product as a set.
Note that you missed $(1,0)$ when you've listed elements from $\mathbb{Z}_2\oplus\mathbb{Z}_4$.
Back to the problem. So you are looking for all subgroups of order $4$. In general there is no easy way to approach this. Fortunately $G$ is a small and not that complicated group.
There are only two (up to isomorphism) groups of order $4$: $\mathbb{Z}_2\oplus\mathbb{Z}_2$ and $\mathbb{Z}_4$.
(1) Let's find all subgroups of $G$ isomorphic to $\mathbb{Z}_4$. Since $\mathbb{Z}_4$ is cyclic then we are looking for elements in $G$ of order $4$. But in the direct product we have a nice formula:
$$o(g,h)=lcm(o(g), o(h))$$
In our situation this means that all we need is an element $h\in\mathbb{Z}_4$ of order $4$ and any element $g\in\mathbb{Z}_2$. Then the cyclic group generated by $(g,h)$ will be a subgroup of order $4$. This gives us following two solutions:
$$H_1=\langle(0,1)\rangle=\{(0,0), (0,1), (0,2), (0,3)\}$$ $$H_2=\langle(1,1)\rangle=\{(0,0), (1,1), (0,2), (1,3)\}$$
Note that there are more elements of order $4$. For example $(1,3)$ but $\langle(1,1)\rangle=\langle(1,3)\rangle$.
(2) Now let's find all subgroups of $G$ isomorphic to $\mathbb{Z}_2\oplus\mathbb{Z}_2$. There is one obvious solution
$$K=\{(0,0), (1,0), (0,2), (1,2)\}$$
Are there any other? The answer is no. That is because any element not from $K$ is of order $4$. Thus no subgroup containing it can be isomorphic to $\mathbb{Z}_2\oplus\mathbb{Z}_2$.
All in all: $\mathbb{Z}_2\oplus\mathbb{Z}_4$ has $3$ subgroups of order $4$.