Denote with $\,{\cal{L}}^G\,$ the space of all continuous functions $\varphi(g)$ mapping the elements of a Lie group $G$ to some vector space ${\cal{L}}$: $$ {\cal{L}}^{G}~=~\left\{~\varphi~~~{\Large{|}}~~~\varphi\,:~G\,\longrightarrow\,{\cal{L}}\right\}~~. $$ Consider a subgroup $K<G$ and its representation $D(K)$ by linear operators on ${\cal{L}}^{G}\,$: $$ D:\quad K~\longrightarrow~{\rm{GL}}({\cal{L}}^{G})~.\qquad\qquad\qquad\qquad\qquad\qquad\qquad (1) $$ It can be induced to the representation $\,\mbox{Ind}_K^GD\,$ on the subspace of functions obeying $$ \varphi(xk^{-1})=D(k)\varphi(x)~.~~~\qquad\qquad\qquad\qquad\qquad\qquad\qquad (2) $$ We now restrict the induced representation to a subgroup $\,B<G\,$. Denoted with $$ U(B)~\equiv~\left(D(K)~\uparrow~G\right)~\downarrow~B~~,\quad\quad\quad\qquad\qquad\qquad\qquad (3) $$ the new representation will be realised with the left translations $$ U(b)~\varphi(g)~=~\varphi(b^{-1}g)~~.~\qquad\qquad\qquad\qquad\qquad\qquad\qquad (4) $$
This representation fixes the double cosets $\,Bu_iK\,$. So the expansion $$ G\,=\,\,\bigcup_i{\cal{U}}_{\,i}\,=\,\,\bigcup_iB\, u_i\, K\,\;\qquad\qquad\qquad\qquad\qquad\quad\quad~ (5) $$ yields a decomposition of $\,U(B)\,$ into subrepresentations: $$ U(b^{\,\prime})\,\varphi(b\, u\, k)~=~\varphi({b^{\,\prime}}^{-\, 1}\, b\,u\, k)~~,\qquad\qquad\qquad\qquad\qquad (6) $$ where $~~b\, u\, k~$ and $~{b^{\,\prime}}^{-\, 1} b\, u\, k~$ belong to the same $~B\, u\, K\,$.
Define the projection operator $\,P_u\,$ so that a function $\,P_u\varphi\,$ coincides with $\,\varphi\,$ on $\,B\, u\, K\,$ and is zero outside it. Then interpret $\,\varphi(g)\,$ as a set of functions $\,\varphi_u\,$ on double cosets: $$ \varphi~=~\left\{\,\varphi_u\,\right\}~~,\qquad\varphi_u\,=~P_u\,\varphi~~.\qquad\qquad\qquad\qquad\quad (8) $$
A function $\,\varphi_u\,$ is fully defined by its values on $\,B\, u\,$, owing to (2) which now becomes $$ \varphi_u(b\, u\, k)=D(k^{-1})\varphi_u(b\, u)~\,.\qquad\qquad\qquad\qquad\qquad~~~ (9) $$ Defining $$ q\,\equiv\, u\, k\, u^{-1}\,\in\, K_u\qquad\qquad\qquad\qquad\qquad (10) $$ and introducing a representation of $\,K_u\,$ $$ D_u(q)\,\equiv\, D(u^{-1}\, q\, u)~~,~~\qquad\qquad\qquad\qquad (11) $$ we can also write the subsidiary condition on $\,\varphi\,$ as a subsidiary condition $$ \Phi_u(b\, q)~=~D^{-1}_u(q)~\Phi_u(b)\qquad\qquad\qquad\qquad (12) $$ on a new function $$ \Phi_u(b\, q)~\equiv\,\varphi_u(b\, q\, u)\qquad\qquad\qquad\qquad\qquad (13) $$ acting as $$ \Phi_u\,:~~~B\,K_u~\longrightarrow~{\cal{L}}~~.~~\qquad\qquad\qquad\qquad (14) $$ Thus, our subrepresentation becomes $\; D_u(K_u)\uparrow B \;$, while the entire representation is $$ \left(D(K)~\uparrow~G\right)~\downarrow~B~=~\int_{B\backslash G/K}du ~D_u(K_u)\uparrow B ~,\qquad\qquad\qquad (15) $$ where the integration goes over all distinct double cosets.
Strangely, this is not what I see in the books. E.g., the Mathematical Encyclopedia says that
https://www.encyclopediaofmath.org/index.php/Induced_representation $$ \left(D(K)~\uparrow~G\right)~\downarrow~B~=~\int_{B\backslash G/K}du ~(D_u(K_u)\downarrow L_u)\uparrow B ~,\qquad~~~ (16) $$ where $$ L_u~\equiv~K_u\cap B~=~u~K~u^{-1}\cap B~~. $$
My question is: why $\,~\downarrow L_u\,$?
This restriction is equivalent to an assumption that in condition (12) we must restrict, by hand, $\,q\,$ from $\,K_u\,$ to $\,L_u\,$. Or to enforce an equivalent restriction of $\,k\,$ in $\,D(K)\,$. Or to restrict the domain of $\,\Phi\,$ to $\,B\,$ in (12). Why should we do that? The desired restriction of the induced representation to $\,B\,$ implies that the left translations must be performed by the elements $\,b\in B\,$. I, however, see no reason why we also should restrict to $\,B\,$ the domain of the functions $\,\Phi\,$ on which this restricted representation acts.
PS.
A possible clue to this problem may lie in the following observation. The subgroup $\,L_u\,$ and the quotient space $\,B/L_u\,$ naturally emerge when we split a double coset $\,BuK\,$ into left cosets $\,\textsf{b}uK\,$.
Indeed, if some $\,b\,$ and $\,b_1\,$ lie in the same left coset, $$ b~u~K~=~b_1~u~K~~, $$ then $$ b~K_u~=~b_1~K_u~~~ $$ or, equivalently, $$ b_1^{-1}b~K_u~=~K_u~~. $$ This obviously entails $\,b_1^{-1}b\,\in\,K_u\,$, wherefrom $$ b^{-1}_1b\,\in\,L_u~~. $$ So the left cosets $\,b\, u\, K\,$ and $\,b_1 u\, K\,$ coincide iff $\,b\,$ and $\,b_1\,$ belong to the same left coset in the quotient space $\,B/L_u\;$,$\,$ with some representative $\,\textsf{b}\,$: $$ b\,,\,b_1\,\in\,\textsf{b}\,{{L}}_{u}\;\,,\;\;\,\textsf{b}\,\in\,B $$ Thus, as many left cosets $\,\textsf{b}\, L_u\,$ in the quotient space $\,B/L_u\,$ so many left cosets $\,X\,=\, \textsf{b}\, u\, K\,$ in the double coset $\,{\cal{U}}=BuK\,$.
We also can say that if $\,b\,$ lies in the left coset $\,b\, u\, K\,$, then $$ b~=~\textsf{b}~l~~,~~~l\in L_u~~, $$ where $\,\textsf{b}\,$ is a representative of the left coset $\,\textsf{b}\,L_u\,$. Thence, $$ \Phi_u(b)~=~D_u^{-1}(l)~\Phi_u(\textsf{b})~~,~~~~~~~~~~~~~~~~~~~~~~~~~~~~\qquad\qquad(17) $$ so the function $\,\Phi_u\,$ is fully defined by its values on a section of the quotient space $\,B/L_u\,$.
This could have answered my question if not for one circumstance: in eqn (12) we still have $\,q\,$ which lies in $\,K_u\,$, not in $\,L_u\,$. So equation (17) becomes $$ \Phi(bq)~=~D_u^{-1}(lq)~\Phi_u(\textsf{b})~~, $$ and the subrepresentation is defined not on $\,L_u\,$ but on $\,K_u\,$ -- unless we fix this ``by hand'' (but on what grounds?)