How to build a finite automaton for right quotient of a regular language?
First of all, note that you are constructing an NFA, and so $\delta'$ should output a set of possible transitions.

You can prove by induction the following identities:
$$
\begin{align*}
\delta'((q,0),w) &amp;= \{ (\delta(q,w),0), (\delta(q,wa),1)\}, \\
\delta'((q,1),w) &amp;= \begin{cases} \{(q,1)\} &amp; w = \epsilon, \\ \emptyset &amp; w \neq \epsilon. \end{cases}
\end{align*}
$$
In particular, $\delta'((q_0,0),w)$ intersects $F \times \{1\}$ iff $\delta(q_0,wa) \in F$. In other words, $w$ is accepted by your NFA iff $wa \in L$, so your automaton is computing $L/a$.

You can simplify your construction by taking the original DFA and simply modifying the set of accepting states, replacing $F$ with
$$
F' = \{ q : \delta(q,a) \in F \}.
$$
The new automaton accepts a word $w$ iff $\delta(q_0,w) \in F'$ iff $\delta(q_0,wa) = \delta(\delta(q_0,w),a) \in F$ iff $wa \in L$.