Consider a group $\,G\,$, a Hilbert space $\,{\mathbb{V}}\,$ with a dot product $\,\langle~,~\rangle\,$, and a space $\,{\cal{L}}^G\,$ of functions $\varphi$ on this group: $$ {\cal{L}}^G\;=\;\left\{~\varphi~\Big{|}~~~\varphi:\,~G\longrightarrow{\mathbb{V}}\,\right\}~~. $$ Let $\,D\,$ be a representation of a subgroup $\,K\leq G\,$ in the said Hilbert space: $$ D~:\quad K~\longrightarrow~GL({\mathbb{V}})\;\;.\qquad\qquad\qquad (1) $$
On a group element $\,g\in G\,$, a function $\,\varphi\in{\cal{L}}^G\,$ assumes the value $\,\varphi(g)\in{\mathbb{V}}\,$. Since this value is a vector in the Hilbert space, we can act on it with some $\,D(k)\,$, $\,k\in K\;$: $$ k\in K~:\quad \varphi(g)\;\mapsto\;D(k)\,\varphi(g)\;\;,\qquad\varphi(g)\in{\mathbb{V}}\;\;. $$ For a fixed $\,g\,$, this is a mapping of one Hilbert-space vector to another.
However, the set of all these mappings, for all $\,g\in G\,$, generates a mapping of a function to a function: $$ k\in K~:\quad \varphi\;\mapsto\;D(k)\,\varphi\;\;,\quad\varphi\in{\cal{L}}^G\;\;. $$ $$ $$ QUESTION 1:
May I write the latter as $$ D~:\quad K~\longrightarrow~GL({\cal{L}}^G)\;\;,\qquad\qquad\qquad (2) $$ using the same notation $\,D\,$ as was used in equation (1)?
REMARK: $~$ While interconnected in an obvious way, the two $\,D$'s are two different representations, because they are acting in different spaces: one in $\,{\mathbb{V}}\,$, another in $\,{\cal{L}}^G\,$. Hence the above question. $$ $$ QUESTION 2:
Would it be possible to say that these two $\,D$'s are, in some sense, equivalent? $$ $$ QUESTION 3:
The induced representation $\,\operatorname{Ind}_K^GD\,$ is implemented with the left translations $$ U_g\varphi(x)=\varphi({g^{-1}}x)~~,\qquad g,\,x\in G\;, $$ acting in the subspace $\,\Gamma\in{\cal{L}}^G$ of the Mackey functions: $$ \Gamma\;=\;\left\{~\varphi~\Big{|}~~~\varphi:\,~G\longrightarrow{\mathbb{V}}\;,\quad \varphi(xk)=D^{-1}(k)\varphi(x)\,\right\}~~. $$ Which of the two $\,D$'s is actually being induced here? -- the $\,D\,$ given by (1) or the $\,D\,$ given by (2)?
NB: for precision, you'd want to say what kind of $V$-valued functions on $G$. Some measurability conditions?
The two representations are not "equivalent" in an immediate sense (though perhaps in a "Morita equivalence" sense...). Also, what is "equivalent" intended to mean?
In any case, contemporary usage would say that the repn of $K$ on $V$ is what is being induced, whatever qualifications or modifiers are put on these function spaces.
EDIT: As in a comment, it is often intended that functions be "square-integrable"... so certainly measurable. When the $K$ here is compact, square-integrability on $G/K$ (or $K\backslash G$...) is equivalent to square-integrability on $G$. Also, for compact $K$, invoking uniform boundedness theorem, a $K$ repn on a Hilbert space can be assumed unitary (without changing the topology on the Hilbert space). So things are as harmonious as they could possibly be.
EDIT2: As in the comment, applying restriction from $G$ to $K$ to the induced repn does not return to the original repn of $K$ at all. In fact, (because the action of $K$ is on the opposite side of $G$, whatever one's convention), that restriction is not even any kind of sum/integral of copies of the original repn of $K$, for non-abelian $G$.