Is it correct to have understanding of logic with the use of set theory, to make learning it easier?

In some sense, the answer to your question is yes. It gives a geometric viewpoint on Boolean algebra (the algebra of propositions), known as Stone duality. A proposition is thought of as some subspace and it can be true or false depending on your position. Even if you don't learn Stone duality, this viewpoint is helpful and sound.

There is indeed a relation between logic expressions and set theory expression. For given set of basic logic sentences $p, q, \dots$ you can create space $\Omega$ of all possible possibilite situations (that is all possible combinations of logical values that your sentences may have). You can relate every logical sentence to the set of situations in which this sentence is true, that is $$ p \text{ is true in situation } x \Leftrightarrow x \in P$$ Then you have for example \begin{align} (p \land q) \text{ is true in situation } x &\Leftrightarrow (p \text{ is true in situation } x )\land (q \text{ is true in situation } x) \Leftrightarrow \\ & \Leftrightarrow (x\in P) \land (x\in Q) \Leftrightarrow \\ & \Leftrightarrow x \in (P \cap Q) \end{align} which means that the set $P \cap Q$ is the set related to the sentence $p \land q$. Analogously you can find the set operations related to other logic operations. The construction is solid, and if you already know set theory it can be useful to learning logic. For example it can visualise various logic laws (like the de Morgan's laws for example) or allow to immediately teanscribe some theorems about set on the theorems about logic sentences. For example, if you know that $$ P \cap (Q \cup R) = (P \cap Q) \cup (P \cap R)$$ then you can immediately tell that sentences $$ p \land (q \lor r) $$ and $$ (p \land q) \lor (p \land r)$$ are equivalent.