This answer addresses the question in the title, not the many all along the post.
The elements of order $2$ of $G:=\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_4$ are:
\begin{alignat}{1}
&a_1=(0,0,2), \space a_2=(0,1,2), \space a_3=(1,0,2), \space a_4=(1,1,2) \\
&a_5=(0,1,0), \space a_6=(1,0,0), \space a_7=(1,1,0) \\
\tag 1
\end{alignat}
Any candidate subgroup of $G$, say $K$, isomorphic to the Klein group must be made of:
- the unit $1_G:=(0,0,0)$;
- any pair $a_i, a_j$;
- the element $a_i+a_j$; in fact: $a_i+a_j\in K$ by closure, and $a_i+a_j\ne 1,a_i,a_j$.
Is, for every $1\le i<j\le 7$, the subset $K_{ij}:=\{1_G,a_i,a_j,a_i+a_j\}$ indeed a subgroup of $G$? It's enough to prove the closure:
- $a_i^2=a_j^2=(a_i+a_j)^2=1_G$
- $a_i+(a_i+a_j)=(a_i+a_i)+a_j=a_j$
- $a_j+(a_i+a_j)=a_j+(a_j+a_i)=(a_j+a_j)+a_i=a_i$
So, indeed $K_{ij}\le G$ and $K_{ij}\cong\mathbb{Z}_2\times\mathbb{Z}_2$.
Now, if we denote $a_k:=a_i+a_j$, then $a_k+a_i=a_j$ and $a_k+a_j=a_i$. So:
- if $k<i<j$, then $K_{ij}=K_{ki}=K_{kj}$;
- if $i<k<j$, then $K_{ij}=K_{ik}=K_{kj}$;
- if $i<j<k$, then $K_{ij}=K_{ik}=K_{jk}$.
Therefore, the number of (distinct) subgroups of $G$ isomorphic to $\mathbb{Z}_2\times\mathbb{Z}_2$ is:
\begin{alignat}{1}
n_K &= \frac{1}{3}\cdot|\{K_{ij}, \space1\le i<j\le 7\}| \\
&= \frac{1}{3}\cdot\frac{7\cdot 7-7}{2} \\
&= 7 \\
\tag 2
\end{alignat}
Explicitly, according to the labelling $(1)$:
\begin{alignat}{1}
K_{12} &= \{1_G,a_1,a_2,a_5\}\space (=K_{15}=K_{25}) \\
K_{13} &= \{1_G,a_1,a_3,a_6\}\space (=K_{16}=K_{36}) \\
K_{14} &= \{1_G,a_1,a_4,a_7\}\space (=K_{17}=K_{47}) \\
K_{23} &= \{1_G,a_2,a_3,a_7\}\space (=K_{27}=K_{37}) \\
K_{24} &= \{1_G,a_2,a_4,a_6\}\space (=K_{26}=K_{46}) \\
K_{34} &= \{1_G,a_3,a_4,a_5\}\space (=K_{35}=K_{45}) \\
K_{56} &= \{1_G,a_5,a_6,a_7\}\space (=K_{57}=K_{67}) \\
\tag 3
\end{alignat}
Indeed, they do have a one-to-one correspondence between, them (in fact, there are $8!$ such correspondences). The problem is that they don't have the "same structure," so none of these correspondences will be homomorphisms, and so no isomorphism exists. But it's tedious in the extreme to check so many functions to see if any of them is a homomorphism, so we instead make use of one of many properties preserved under isomorphisms. Two such (related) properties are that:
- isomorphisms preserve orders of elements (meaning that any given element will be mapped to an element of the same order), and
- isomorphic images of cyclic groups are again cyclic groups.
Those two properties are probably the simplest way to show that these two groups are not isomorphic, since (in particular) there are four different order $8$ elements of $\Bbb Z_8,$ but no such elements in $\Bbb Z_2\times\Bbb Z_4.$ So, we can directly conclude that they are not isomorphic by fact 1. Alternately, we can note that this means $\Bbb Z_8$ is cyclic and $\Bbb Z_2\times\Bbb Z_4$ is not, so can indirectly conclude that they are not isomorphic by fact 2.
There are many other such properties that we can use in other cases. It's good to keep several of them in mind. As Mark points out in the comments, though, the first property is a necessary condition for a function to be an isomorphism, but not sufficient. However, two cyclic groups of the same order will necessarily be isomorphic. Unfortunately, most groups aren't cyclic, so this property won't usually be useful.
Best Answer
Hint: can you find an element of order $4$ in any of these groups?