Let $\phi \colon H \to \mathbb{R}^d$ be a linear map defined on a real Hilbert space $H$.
Let $e_i$ denote the standard basis, and $\phi_i = e_i^\ast \circ \phi$ denotes the projection of $\phi$ onto the $i$th coordinate.
By the Riesz representation theorem, if the following quantity is finite $$\sup_{\|x\|_H = 1} |\phi_i(x)| = \sup_{\|x\|_H = 1} |\langle e_i, \phi(x) \rangle|, $$ then there exists a vector $y_i \in H$ such that $\phi_i(x) = \langle y_i, x\rangle_H$.
Thus, if the following is finite (note using a single $x$ below only works since $d$ is finite) $$ \sup_{\|x\|_H = 1} \|\phi(x)\|_\infty = \sup_{\|x\|_H = 1} \max_i |\langle e_i, \phi(x) \rangle|, $$ then $\phi(x) = \sum_{j=1}^d \langle y_j, x\rangle_H e_j$, for some sequence $\{y_j\}_{j=1}^d$ in $H$.
That is, we can define the following operator $$ T = \sum_{j=1}^d e_j \otimes y_j $$ and it has the property that $\phi(x) = Tx$.
Is there a generalization of this type of vector-valued version of Riesz theorem when the output of $\phi$ is infinite dimensional?
No. An operator of the form $T = \sum_{j=1}^d e_j \otimes y_j$ is finite-rank. A limit of such operators is compact. That is, even if you allow $d=\infty$ in the above equality and the series converges, all you get is a compact operator.
As a over-arching idea, there is no "general form" for a bounded operator between two infinite-dimensional Hilbert spaces.