Consider the space $\,{\cal{L}}^G\,$ of all continuous functions $\,G\longrightarrow{\cal{L}}\,$ mapping a Lie group $\,G\,$ into a vector space $\,{\cal{L}}\,$.
Assume that $\,G\,$ has two proper subgroups: $$ K\,,~Q~<~G~~, $$ whose representations, $\,D(K)\,$ and $\,\Lambda(Q)\,$, are acting in $\,{\cal{L}}^G\,$.
Consider two subspaces of $\,{\cal{L}}^G\,$. One subspace,
$$
{\mbox{Map}}_K(G,\,{\cal{L}})\,=\,\left\{\,\varphi\,\right\}~~,
$$
comprises the vector functions $\,\varphi\,$ obeying the equivariance condition
$$
\varphi(g\, k)~=~D^{-1}(k)~\varphi(g)~,~~~k\,\in\, K~~.
$$
In this subspace, $\,D(K)\,$ is induced to a representation of $\,G\,$, denoted by
$$
U^{(D)}\,\equiv\,D(K)\,\uparrow\, G
$$
and implemented with
$$
U^{(D)}_g\,\varphi(g^{\,\prime})~=~\varphi(g^{-1}\, g^{\,\prime})~~.
$$
Another subspace, $$ {\mbox{Map}}_Q(G,\,{\cal{L}})\,=\,\left\{\,\psi\,\right\} $$ will comprise the functions $\,\psi\,$ satisfying $$ \psi(g\, q)~=~\Lambda^{-1}(q)~\psi(g)~,~~~q\,\in\, Q~~. $$ In this subspace, $\,\Lambda(Q)\,$ is induced to a representation of $\,G\,$, denoted by $$ U^{(\Lambda)}\,\equiv \,\Lambda(Q)\,\uparrow\, G $$ and implemented with $$ U^{(\Lambda)}_g\,\psi(g^{\,\prime})~=~\psi(g^{- 1}\, g^{\,\prime})~~. $$
While both $\,U^{(D)}\,$ and $\,U^{(\Lambda)}\,$ are realised via left translations, they are different representations, as they are acting in subspaces defined by different subsidiary conditions.
For convenience, we summarise this in the table: $$ \varphi(g\, k)=D^{-1}(k)\,\varphi(g)~,~~k\in K \qquad \quad \psi(g\, q)=\Lambda^{-1}(q)\,\psi(g)~,~~q\in Q $$ $$ U^{(D)}\,\equiv\,D(K)\,\uparrow\, G \qquad \qquad \qquad \qquad U^{(\Lambda)}\,\equiv \,\Lambda(Q)\,\uparrow\, G $$ $$ U^{(D)}_g\,\varphi(g^{\,\prime})~=~\varphi(g^{-1}\, g^{\,\prime}) \qquad \qquad \qquad \qquad U^{(\Lambda)}_g\,\psi(g^{\,\prime})~=~\psi(g^{- 1}\, g^{\,\prime})~ $$
Our goal is to describe the space $\,\left[\, D(K)\,\uparrow\, G\,,~\Lambda(Q)\,\uparrow\, G \,\right]\,$ of the morphisms $\,\psi\,=\,\hat{T}\,\varphi\,$.
QUESTION:
How to prove that the most general form of a morphism is $$ \psi(g)~=~(\hat{T}\,\phi)(g)~=~\int_G t(g^{-1}\, g^{\,\prime})\,\varphi(g^{\,\prime})\,dg^{\,\prime}~~,\qquad\qquad\qquad(1) $$ where $\,dg\,$ is an invariant measure on $\,G\,$.
PS.
As an aside, I would mention that for the equivariance conditions to be satisfied the kernel must obey $$ t(qgk) = \Lambda(q) t(g) D(k)~~,~~~q\in Q\,,~~g\in G\,,~~k\in K~~. $$ This, however, is the next theorem; and I don't want to go there until the basic property (1) is proven.
I have now found a remarkably simple proof of the fact that the variables $\,g\,$ and $\,g^{\,\prime}\,$ enter $\,t(g\,,\, g^{\,\prime})\,$ through convolution: $$ t(g\,,\, g^{\,\prime})\;=\;t(g^{-1}\, g^{\,\prime})~~.\qquad\qquad\qquad (*) $$ To prove this, we should use the definition of an intertwiner, $$ U_{g_1}^{(A)}\,T\,\varphi~=~T\,U^{(D)}_{g_2}\,\varphi~~.\qquad\qquad\qquad (**) $$ Its left-hand side can be written down as $$ (\, U_{g_1}^{(A)}\, T\,\varphi)\,(g_2)\,=\,(T\,\varphi)\,(g_1^{-1}g_2) \,=\,\int t(g_1^{-1}g_2\,,\,g^{\,\prime})\,\varphi(g^{\,\prime})\,dg^{\,\prime}\,~. $$ Its right-hand side can be processed as $$ (T\,U_{g_1}^{(D)}\,\varphi)\,(g_2)$$ $$=\,\int t(g_2\,,\,g^{\,\prime\prime})\,(U_{g_1}^{(D)}\varphi)\,(g^{\,\prime\prime})\,dg^{\,\prime\prime} \,=\,\int t(g_2\,,\,g^{\,\prime\prime})\,\varphi(g^{-1}\, g^{\,\prime\prime})\,dg^{\,\prime\prime} $$ $$ =\int t(g_2,\,g_1(g_1^{-1}\, g^{\prime\prime}))\,\varphi(g^{-1}g^{\prime\prime})\,dg^{\prime\prime} =\int t(g_2,\,g_1 g^{\prime})\,\varphi(g^{\prime})\, dg^{\prime}~~, $$ where we defined $\,g^{\,\prime}=\, g_1^{-1}\, g^{\,\prime\prime}\,$ and assumed the measure left-invariant, $\,dg^{\,\prime}=\, d(g_1^{-1} g^{\,\prime\prime}) \,=\, dg^{\,\prime\prime}\,$.
If we plug the right-hand sides of the former and latter formulae in equation (**), we shall arrive at $$ t(g_1^{-1}g_2\,,\,g^{\,\prime})~=~t(g_2\,,\,g_1 g^{\,\prime})~~. $$ If we now set $\,g_2=1\,$ and $\,g=g_1^{-1}\,$, the above formula will shape into $$ t(g\,,\,g^{\,\prime})~=~t(1\,,\,g^{-1} g^{\,\prime})~~. $$ We now can rename $\,t(1\,,\,g^{-1} g^{\,\prime})\,$ as $\,t(g^{-1} g^{\,\prime})\,$ and regard (*) proven: $\,t(g\,,\, g^{\,\prime})\,=\, t(g^{-1}\, g^{\,\prime})\,$. Accordingly, $$ \psi(g)\,=\,(\hat{T}\,\phi)(g)\,=\,\int t(g^{-1}\,g^{\,\prime})\,\varphi(g^{\,\prime})\,dg^{\,\prime}\;\;. $$ My understanding is that this integral may, in principle, diverge even when the representations are defined on nuclear functions. (Please correct me if I am wrong.) In such situations, one has to establish analytic continuation, i.e. to define this integral in the complex domain, where is does converge to an analytic function.