Constructive proof to show the quotient of two regular languages is regular
The quotient of two languages $R$ and $L$ is
$$R/L = \left\{x\in \Sigma^{*} : \exists y \in L. xy \in R\right\}$$

Here is a constructive solution using a non-deterministic finite automaton (NFA) to show $R/L$ is regular when both $R$ and $L$ are regular.

Let $(Q_1,\Sigma,\delta_1,s_1,F_1)$ and $(Q_2,\Sigma,\delta_2,s_2,F_2)$ be deterministic finite automata for $R$ and $L$ respectively. Define a NFA

$$D=(Q_1 \times \{s_2\} \times \{1\} \cup Q_1 \times Q_2\times \{2\}, \Sigma, \delta, (s_1,s_2, 1), F_1 \times F_2\times \{2\})$$
where the transitions are:


$\delta((q_1,s_2,1),\sigma) = \{ (\delta_1(q_1,\sigma),s_2, 1)\}$.
$\delta((q_1,s_2,1),\epsilon) = \{ (q_1,s_2, 2)\}$.
$\delta((q_1,q_2,2),\sigma) = \{ (\delta_1(q_1,\sigma),\delta_2(q_2,\sigma), 2)\}$.


I'll leave it as an exercise for you to verify the language accepted by $D$ is $R/L$.