Let $G$ be a compact group and $V$ a finite-dimensional vector space with a continuous $G$-action. Consider a linear map $\phi: V \to L^2(G)$ satisfying that for any $v \in V, h \in G$:
$$ \phi(v)(g h) = \phi(h \cdot v)(g) \quad \text{for almost all $g \in G$} $$ Must $\phi(v)$ be continuous for any $v$?
This is used (implicitly) in Serganova's A Journey Through Representation Theory (Chapter 2, Lemma 2.3) to prove that matrix coefficients $$ \begin{align} V^* \times V \times G &\to \mathbb{C} \\ \alpha, v, g &\mapsto \alpha(g \cdot v) \end{align} $$ provide an isomorphism $V^* \cong \operatorname{Hom}_G(V, L^2(G))$, and ultimately prove the Peter-Weyl theorem.
The answer is yes, $\phi(v)$ is equal to a continuous function a.e. for all $v$.
Here is a proof.
The assumption about $\phi$ precisely says that it is a covariant map from $V$ to $L^2(G)$, where $L^2(G)$ is equipped with the right regular representation of $G$.
Therefore the range of $\phi$, henceforth denoted $R$, is an invariant subspace.
Splitting $R$ as a direct sum of irreducible subspaces we may assume WLOG that $R$ itself is irreducible.
By the Peter Weyl Theorem (https://en.m.wikipedia.org/wiki/Peter%E2%80%93Weyl_theorem) we have that $R$ is generated by the matrix coefficients in $R$, which are known to be continuous. Therefore $R$ is formed by continuous functions.
EDIT: Let me expand a bit on the assertion above.
The conceptual reason is the uniqueness of the decomposition of a unitary representation into irreducible ones. To better explain this let $\{\pi _i:i\in I\}$ be a set of representatives for the equivalence classes of irreducible representations of $G$.
Decomposing $R^\perp$ into irreducibles, say $\bigoplus_{i\in I} n_iH_{\pi _i}$, we have that $$ L^2(G) = R\oplus \bigoplus_{i\in I} n_iH_{\pi _i}, $$ at the same time that we have the standard decomposition of $L^2(G)$ given by the Peter Weyl Theorem $$ L^2(G) = \bigoplus_{i\in I} d(\pi _i)H_{\pi _i}. $$ Thus, if $\pi _{i_0}$ is the class of the representation of $G$ on $R$, we have that $$ R\oplus n_{i_0}H_{\pi _{i_0}} = d(\pi _{i_0})H_{\pi _{i_0}}, $$ by uniqueness, and in particular $ R\subseteq d(\pi _{i_0})H_{\pi _{i_0}}. $ Since $d(\pi _{i_0})H_{\pi _{i_0}}$ is spanned by the (continuous) matrix coefficients associated to $\pi _{i_0}$, we conclude that $R$ is formed by continuous functions.
A more pedestrian approach is as follows: denote by $\rho $ the representation of $G$ on $R$, and let $\{e_i\}_{1\leq i\leq n}$ be an orthonormal basis for $R$.
We will show that each $e_i$ is orthogonal to every matrix coefficient associated to any irreducible representation $\pi $ not equivalent to $\rho $.
By Peter-Weyl we will then deduce that each $e_i$ is a finite linear combination of matrix coefficients associated to $\rho $, which are continuous functions, thus proving the $e_i$ to be continuous.
Observe that since the regular representation restricts to $\rho $ on $R$, for every $g$ and $h$ in $G$, we have that $$ e_j(hg) = \rho_g e_j(h) =\sum_{i=1}^n u_{ij}(g)e_i(h), $$ where the $u_{ij}$ are the matrix coefficients of $\rho $ in the given basis.
Let $\pi $ be another irreducible representation of $G$ which is inequivalent to $\rho $, and let $x$ and $y$ be vectors in the space of $\pi $, so that $$ c(g) := \langle x, \pi _{g}(y)\rangle $$ defines a matrix coefficient for $\pi $. As we already said, we will next prove that $c$ is orthogonal to each $e_j$.
By invariance of the Haar measure we have for every $g$ that $$ \langle c, e_j\rangle = \int_G \overline{c(h)} e_j(h)\, dh = \int_G \overline{c(hg)} e_j(hg)\, dh = $$$$ = \sum_{i=1}^n \int_G \overline{ c(hg) } u_{ij}(g)e_i(h)\, dh = \cdots $$ Observing that this does not depend on $g$, we may integrate it against $g$. After doing so and exchanging the order of integration we see that the above equals $$ \cdots = \sum_{i=1}^n \int_G e_i(h)\left(\int_G \overline{ c(hg) } u_{ij}(g)\, dg\right) \, dh. $$ Notice that the term within parenthesis is the inner-product in $L^2(G)$ of the matrix coefficient $$ g\mapsto c(hg) = \langle x, \pi _{hg}(y)\rangle = \langle \pi _{h^{-1}}(x), \pi _{g}(y)\rangle $$ by the matrix coefficient $u_{ij}$, so it vanishes by the Peter-Weyl orthogonality relations since $\pi $ and $\rho $ are inequivalent.