One way to answer this is given by what is called Frobenius reciprocity (for simplicity, I am assuming that we are working over the complex numbers here).
This states that the constituents of the induced representation are precisely those which, when restricted to the subgroup, have the trivial representation as a constituent.
The above also works for any other irreducible representation of the subgroup, and even says what the multiplicities are (being the same as the multiplicity when one restricts).
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.
Best Answer
Frobenius Reciprocity says that induction and restriction are adjoint functors. If you aren't comfortable with that language, what this means is that if $V$ is a representation of $H$ and $W$ is a representation of $G$ then there is an isomorphism:
$$Hom_G(Ind_H^G (V), W) \cong Hom_H(V, Res^G_H (W)) $$
Really adjunction gives something stronger, a sort of consistent way to choose such an isomorphism for all such pairs $V$ and $W$, but for a lot of purposes this statement is what we really care about.
Applying this to your question, where $V = 1_H$ and $W = 1_G$, we get:
$$Hom_G(Ind_H^G (1_H), 1_G) \cong Hom_H(1_H, Res^G_H (1_G)) $$
The point is that while the left hand side involves a term $Ind_H^G (1_H)$ we don't quite understand, the left hand side is easy since clearly $Res^G_H (1_G) = 1_H$. So this just becomes
$$Hom_G(Ind_H^G (1_H), 1_G) \cong Hom_H(1_H, 1_H) \cong \mathbb{C}$$
So this tells us that $Hom_G(Ind_H^G (1_H), 1_G)$ is one dimensional, which exactly means that there is a single copy of the trivial representation of $G$ in $Ind_H^G (1_H)$, since if there were two copies we could project onto each one and scale them independently giving us a(n at least) two dimensional Hom-space.