Question: "I am studying representations of Lie groups and I still cannot find the intuition behind Frobenius Reciprocity theorem."
Answer: Let $i:H \rightarrow G$ be an inclusion of arbitrary groups and let $W$ be a left $H$-module and $U$ a left $G$ module (let $W,U$ be $k$-vector spaces with $k$ a field). If you consider the group algebras $k[H], k[G]$ and view $W,V$ as modules over the group algebras, the above mentioned FR says there is a natural isomorphism
$$F:Hom_{k[H]}(W, U_{k[H]}) \cong Hom_{k[G]}(k[G]\otimes_{k[H]} W, U).$$
with inverse $F'$.
Given any map of $k[H]$-modules $\phi: W \rightarrow U_{k[H]}$ you get an induced map
$$F(\phi):k[G]\otimes W \rightarrow U$$
defined by $F(\phi)(x\otimes w)=x\phi(w) \in U$. Conversely given a map of $k[G]$-modules
$$\psi: k[G]\otimes W \rightarrow U$$
there is an induced map
$$F'(\psi): W \rightarrow U$$
defined by
$$F'(\psi)(w):=\psi(1\otimes w).$$
You may check that $F,F'$ are inverses of each other and that the above is an isomorphism for any $H,G,W,U$. With this formulation, the FR theorem becomes a statement about Hom and tensor products of modules over associative rings.
The proof is straight forward: You must verify that the maps $F,F'$ defined above satisfy $F \circ F' = F' \circ F = Identity$ - their compositions are the "identity map".
Note: You can formulate a similar statement in terms of the universal enveloping algebras $U(Lie(H)), U(Lie(G))$ of the Lie groups $H,G$: There is a "natural isomorphism"
$$F:Hom_{U(Lie(H))}(W, U_{U(Lie(H))}) \cong Hom_{U(Lie(G))}(U(Lie(G)\otimes_{U(Lie(H))} W, U).$$
In fact for any map of associative $k$-algebras $\rho: A \rightarrow B$ with $W\in Mod(A), V\in Mod(B)$, there is a "natural isomorphism"
$$F(A;B): \text{ } Hom_A(W,V_A) \cong Hom_B(B\otimes_A W, V).$$
Frobenius reciprocity for $H \subseteq G$ is a special case of this isomorphism $F(A,B)$ - it is a statement about a relation between $Hom$ and $\otimes$ for modules over associative rings. When you take "derived functors" you get similar relations.
