Proving the L' = {w|deleting one symbol from a string in L} is regular when L is regular
Maybe the formal proof below will help you to understand the construction using a NFA.
Let $\mathcal{A} = (Q, A, \cdot, i, F)$ be a deterministic automaton recognising your language. For each pair of states $(p, q)$, let $L_{p,q}$ be the language recognised by the automaton $(Q, A, \cdot, p, \{q\})$. Thus $L_{p,q}$ is the set of labels of all paths from $p$ to $q$. By construction, each language $L_{p,q}$ is regular. Let $T$ be the set of transitions of $\mathcal{A}$, that is,
$T = \{ (p,a,q) \in Q \times A \times Q \mid p \cdot a = q \}.$
Let
$$
R = \bigcup_{(p,a,q) \in T}  \bigcup_{f \in F} L_{i,p}L_{q,f}
$$

Claim. $D(L) = R$.

Indeed, let $w \in D(L)$. By definition, there exist some words $u$ and $v$ and a letter $a$ such that $w = uv$ and $uav \in L$. Let $p = i\cdot u$, $q = p \cdot a$ and $f = q \cdot v$. By construction, $u \in L_{i,p}$, $v \in L_{q,f}$, $(p,a,q) \in T$ and $f \in F$. Thus $w \in R$.
Let now $w \in R$. Then there exist some states $p, q \in Q$ and $f \in F$  and some transition $(p,a,q) \in T$, such that $w \in L_{i,p}L_{q,f}$. It follows that $w = uv$ for some $u \in L_{i,p}$, and $v \in L_{q,f}$. Now $i \xrightarrow{u} p \xrightarrow{a} q \xrightarrow{v} f \in F$, which means that $uav \in L$. Consequently, $w \in D(L)$, which proves the claim.
Coming back to your question, the first copy of $\cal A$ in ${\cal A}'$ is used to accept $L_{i,p}$ and the second one to accept $L_{q,f}$. The $\epsilon$-transitions are used to pass from the first copy to the second one.