Operator-valued vectors and representation theory

Let us not be so abstract as to lose all physical terminology. What you're talking about here is a consistency condition for fields being quantized.

A classical field $A^\mu$ transforms in some finite-dimensional representation $\pi: V \to V$ of some symmetry group $G$ (e.g. the Lorentz group) as $(\pi(g)\phi)^\mu = \pi(g)^\mu_\nu \phi^\nu$.

In quantum physics, this symmetry group gets represented on a Hilbert space as $U: H \to H$. Since the components of the quantum field are promoted to operators, they transform like all operators do under this transformation: $A^\mu \mapsto U^\dagger(g)A^\mu U(g)$.

Eq. (5) is a condition that "quantization" commutes with "apply $G$", i.e. the result of doing the transformation on the classical field and then interpreting it as an operator is the same as transforming the field operators themselves.

If you want to state rigorously in which space eq. (5) holds, then this is simply the tensor product $V\otimes \mathcal{O}(H)$, where $\mathcal{O}(H)$ is some suitable space of operators on $H$ that contains the quantized $\phi^\mu$.

The situation you describe exactly arises when representing a finite dimensional real Lie group $G$ with a strongly continuous representation (finite or infinite dimensional) $$G\ni g \mapsto U_g\:.\tag{1}$$ In this case one focus on the natural representation the group has on its Lie algebra: $$L_g : \mathfrak{g} \to \mathfrak{g}$$ defined fruom the requirement $$g e^{t s} g^{-1} = e^{tL_gs}$$ (taking the derivative for $t=0$ of both sides).

The representation $G\ni g \mapsto L_g$ is finite dimensional since the Lie algebra is finite dimensional, thus it is constructed by matrices. $$L_g s_k = \sum_r A_k^r(g) s_r$$ for every basis $s_1,\ldots, s_n$ of the Lie algebra $\mathfrak{g}$.

From known theorems on strongly continuous unitary representations, the unitary operators in (1) can be written, in a neighborhood of the identity, $$U_g = \exp(-i \sum_r a^k S_k)$$ where all linear combinations of the operators $S_k$ generally defined in an infinite dimensional Hilbert space where (1) exists, are essentially self adjoint on a common invariant (also for the representation $U$) dense domain and the map $dU: s_k \mapsto -iS_k$ induces a Lie algebra representation on that space. This domain is known as the Garding domain (another similar domain is that discovered by Nelson).

$S_k$ is here the generator, in the sense of the Stone theorem, of the one parameter group $U_{e^{ts_k}}$.

It is now easy to prove that $$U_g S_k U_g^{-1} = \sum_r A_k^r(g) S_r$$ is true on the said domain.

What you call operator valued elements are nothing but the elements in the image of a Lie algebra representation $dU$ of the Lie algebra $\mathfrak{g}$ induced by a strongly continuos unitary representation of the Lie group $G$.