Given that $G$ is subdirect product of $A$ and $B,$ and $G/G'$ is not subdirect product of $A/A'$ and $ B/B',$ this group contains $G'$? In this paper, the following definitions are given:

Recall that if  $AB=A\times B$ is a direct product of groups $A$ and $B$, a subgroup $G$ of $AB$ is called a subdirect product of $A$ and $B$ if $AG=BG=AB$.
Let $G$ be a subdirect product of two groups, $A$ and $B$, such that $G/G'$ is not a subdirect product of $A/A'$ and $B/B'$, and set
$$R=G'(B\cap G)\cap G'(A\cap G).$$

Shortly after, the following claim is made:

As $G/G'$ is not a subdirect product of $A/A'$ and $B/B'$, it is easy to see that $R > G' > 1$.

As it is easy to see I suspect I am overlooking something obvious, but why is $R>G'>1$?
 A: Let $G$ be a subdirect product of two groups, $A$ and $B$, such that
$G/G′$ is not a subdirect product of $A/A′$ and $B/B′$, and set
$R=G′(B\cap G)\cap G′(A\cap G)$. [...]
As G/G′ is not a subdirect product of A/A′ and B/B′, it is easy to see that
$R\gt G′ \gt 1$.
The statement is not correct, since $G/G'$ is a group
consisting of equivalence classes
of pairs, while any subdirect product of $A/A'$ and $B/B'$ is a group
consisting of pairs of equivalence classes. These are not the same kinds
of objects.
Instead of saying
``$G/G′$ is not a subdirect product of $A/A′$ and $B/B′$'',
what the author probably should have said is that the canonical map
of $G$ to $(A/A′)\times (B/B′)$ has kernel properly containing $G'$.
Equivalently, the restriction to $G$ of $A'\times B'=AB'\cap A'B=
AG'\cap BG'$
properly contains $G'$. In symbols, this is $G\cap(AG'\cap BG') \gt G'$. This is a statement about subgroups of $A\times B$, and $A$ is being identified with $A\times \{1\}$, $B$ with $\{1\}\times B$, and $G$ is being considered to be a subgroup of $A\times B$.
But $G\cap(AG'\cap BG') \gt G'$ is basically what you are being asked to show,
since it can be shown that $R=G\cap(AG'\cap BG')$.
This claim relies on the facts that
$G'(A\cap G)=G\cap(AG')$ and 
$G'(B\cap G)=G\cap(BG')$, for then
$R=G'(A\cap G)\cap G'(B\cap G) = G\cap (AG'\cap BG')$.
So it is necessary to explain why, say, 
$G'(A\cap G)=G\cap(AG')$. Let $P=G'(A\cap G)$ 
represent the left hand side
and let $Q=G\cap(AG')$ represent the right hand side.
Since $G'\subseteq Q$ and $(A\cap G)\subseteq Q$,
we have $P\subseteq Q$. Conversely, if $x\in Q$, then
$x\in G$ and $x=ag'$ for some $g'\in G'$ and some $a\in A$.
Then $a=x(g')^{-1}\in G$, so $a\in A\cap G$. Hence
$x=ag'\in (A\cap G)G' = P$. This shows that $P=Q$.
A: Partial answer:
This is a partial answer I came up with when I tried answering this question myself. I got stuck on my last step and don't have time to keep working on it, which is why I placed a bounty.  I then figured my partial answer could potentially be of help to someone trying to answer this completely, so here you go. 
Preliminaries: 
First, let me clarify/recall some notation and properties:


*

*In the question (and this answer, for consistency), many groups are identified with a factor of a direct product, e.g. $A \cong (A \times 1)$ and $B \cong (1 \times B)$, hence the notation $AB = (A \times 1)(1 \times B) = A \times B$. 

*$G_1 > G_2$ means $G_2$ is a proper subgroup of $G_1$.

*Another equivalent definition of a subdirect product $G \leq A \times B$: $$\forall a \in A: \exists b \in B: (a,b) \in G \;\;\;\text{ and }\;\;\;\forall b \in B: \exists a \in A: (a,b) \in G.$$
Proof:
Let's first verify that $R \geq G' \geq 1$, without bothering if the subgroups are proper or not. The second "inequality" holds trivially, so we only have to bother with the first one:

Proof: Let $g \in G'$. Since $1 \in (B \cap G)$, we have that $g = g \cdot 1 \in G' (B \cap G)$. Similarly we have that $g \in G'(A \cap G)$. Hence $$g \in  G' (B \cap G)  \cap G'(A \cap G) = R.$$

Now, onwards to the "proper" part. We first show that $G' > 1$, i.e. $G' \neq 1$.

Lemma: If $G' = 1$, then $A' = B' = 1$.
Proof: Suppose (WLOG) that $A' \neq 1$, i.e. $\exists a_1,a_2 \in A$ such that $[a_1,a_2] \neq 1$. Because $G$ is a subdirect product of $A \times B$, we find that $\exists b_1,b_2 \in B$ such that $(a_1,b_1),(a_2,b_2) \in G$. 
But then $1 \neq  ([a_1,a_2],[b_1,b_2]) = [(a_1,b_1),(a_2,b_2)] \in G'$, so $G' \neq 1$. The same reasoning holds if $B' \neq 1$, so by contraposition this proves the lemma.

We can now prove that $G' > 1$ by contradiction:

Proof: Suppose that $G' = 1$. Then $A'= B' = 1$, by the previous lemma, hence $G/G' = G$, $A/A' = A$ and $B/B' = B$. But then clearly $G/G'$ is a subdirect product of $A/A'$ and $B/B'$: contradiction!

Finally, we have to show that $R > G'$. 
This is the part I can't solve. Some thoughts (which may or may not be correct): 


*

*If $x \in R$, then $\exists g,g' \in G,a \in A,b \in B$ such that
$$ x = g(a,1) = g'(1,b).$$
But then $(a,b^{-1}) = g^{-1}g' \in G' \leq A' \times B'$, so we may assume that $a' \in A'$ and $b' \in B'$.

*If $G' = A' \times B'$, then $G/G'$ is a subdirect product of $A/A' \times B/B'$. If not, then I don't know how $G/G'$ is identified with a subgroup of $A/A' \times B/B'$ at all.
