Suppose the domain of the functions of $\mathcal H$ is $X$. Let $k: X\times X\rightarrow \mathbf R$ be the reproducing kernel, which is symmetric and positive semidefinite. Define
$$L[f] := \sum_{i=1}^n \big(y_i -\beta_i^
T\vec f\big)^2 + \lambda \|f\|^2_\mathcal{H}, \quad\lambda\ge0$$
where $\vec f := \big[f(x_1),f(x_2),\cdots,f(x_m)\big]^T$ and $\beta_i$ is a column matrix. Let $M:=
\{1,2,\cdots,m\}$ and $N:=
\{1,2,\cdots,n\}$.
(The factor $\frac1n$ in the original question formulation is absorbed into $\lambda$ so as to abbreviate the notations.)
$\forall f\in\mathcal H,$
$$f=\sum_{i=1}^m c_i k(\cdot,x_i)+v$$
for some $(c_i)_{i\in M}$ and some $(v\in\mathcal H) \perp \operatorname{span}(\{K(\cdot,x_i)|i\in M\})$ or $\langle v,k(\cdot,x_i)\rangle=0,\,\forall i\in M$. So for $\forall j\in M$
\begin{align}
f(x_j)&=\langle f,k(\cdot,x_j) \rangle \\
&=\sum_{i=1}^m c_i \langle k(\cdot,x_i),k(\cdot,x_j)\rangle +\langle v,k(\cdot,x_j)\rangle \\
&= \sum_{i=1}^m c_i k(x_i,x_j),
\end{align}
and $f(x_j)$ is independent of $v$.
\begin{align}
\|f\|^2_{\mathcal H}&=\Big\langle \sum_{i=1}^m c_i k(\cdot,x_i), \sum_{j=1}^m c_i k(\cdot,x_j) \Big\rangle+\|v\|_\mathcal H^2 \\
&= \sum_{i,j=1}^m c_i k(x_i,x_j) c_j+\|v\|^2_\mathcal H
\end{align}
since $v\perp$ span$(k(\cdot,x_i)|i\in M)$.
Therefore the $f$ that minimizes $L[f]$ should have $v=0$. We denote such $f$ by $f^*$, and
$$f^*:=\operatorname{arg min}_{f \in\mathcal H}L[f]=\sum_{i=1}^m c_i k(\cdot,x_i)$$
and
$$f^*(x_j)=\sum_{i=1}^m c_i k(x_j,x_i),\quad\forall i\in M,$$
or in the matrix form
$$\vec {f^*}=Kc$$
where $m\times m$ matrix $K$ has $(j,i)$'th entry $K_{j,i}:=k(x_j,x_i),\,\forall i,j \in M$ and $m\times 1$ matrix $c$ has entry $c_i,\,\forall i\in M$. Matrix $K$ is symmetric since the function $k(\cdot,\cdot)$ is symmetric.
Let the $n\times 1$ matrix $y:=[y_i]_{i\in N}$, and $n\times m$ matrix $\beta:= [\beta_i^T]_{i\in N}$ be the matrices constructed from stacking the components on the right hand side vertically. Substituting $f=f^*$ into $L[f]$, we have
\begin{align}
L[f^*]&=(y-\beta Kc)^T(y-\beta Kc)+\lambda c^T Kc \\
&= c^T(K\beta^T\beta K+\lambda K)c-2y^T\beta K c +y^Ty.
\end{align}
Thus $L[f^*]$ minimizes at
$$0=\big(K\beta^T\beta K+\lambda K\big)c-K \beta^T y=Ku,$$
with
$$u:=(\beta^T\beta K+\lambda I)c-\beta^T y.$$
It is true iff $u$ is in the null space of $K$ or that $u$ is a zero eigenvector $K$. If matrix $K$ is positive definite, the left $K$'s on both sides of the equation cancels and we have
$$c= (\beta^T\beta K+\lambda I)^{-1}(\beta^Ty+u)$$
so long as $\lambda>0$ or $\beta$ is of full rank. If $K$ is positive definite, $u=0$.
Remark:
The kernel function $k(⋅,⋅)$ is positive definite. That however does not guarantee the positive definiteness of matrix $K$. For example, the $K$ resulting from $\vec x$ with duplicating $x_i$ is not positive definite. All distinct $x_i$'s would not guarantee that either.
$\{k(\cdot,x_i)\}_{i\in M}$ with $x_i$'s all distinct is not necessarily linearly independent. For example, the linear kernel $k(x,y)=x^Ty,\, \forall x,y\in R^p$ for some natural number $p$. $\{k(\cdot,x_i)\}_{i\in M}$ is linearly dependent when $\{x_i\}_{i\in M}$ is.
