The vertices are the corners of the triangle. Suppose $H = \{v_1, v_2, v_3\}$, and assume that the points are not collinear (otherwise the uniqueness conclusion is false).
Then $K = \text{co} \{v_i\}_i$, and hence $x \in K$ iff $x = \sum_i \lambda_i v_k$ with $\sum_i \lambda_i = 1$ and $\lambda_i \geq 0$. Furthermore, since the points of $H$ is not collinear, the $\lambda_i$ are unique. The $\lambda_i$ are known as the barycentric coordinates of $x$ with respect to $\{v_i\}_i$.
Also, note that each $f \in A$ is affine, that is, if $\sum \lambda_i = 1$, then $\sum_i \lambda_i f(v_i) = f(\sum_i \lambda_i v_i)$.
To show existence, define the measure $\mu_x A = \sum_i \lambda_i 1_A(v_i)$, where the $\lambda_i$ are the barycentric coordinates of $x$. Then if $f \in A$, we have $\int f d \mu_x = \sum_i \lambda_i f(v_i) = f(\sum_i \lambda_i v_i) = f(x)$.
To show uniqueness, let $x \in K$ and suppose $\mu$ is a measure supported on $H$ such that $f(x) = \int_H f d \mu$ for all $f \in A$. This means $\int_H f d \mu = \int_H f d \mu_x$ for all $f \in A$. Since $\mu$ is supported on $H$, it has the form $\mu A = \sum_i \mu\{v_i\} 1_A(v_i)$, so we now have $\sum_i \lambda_i f(v_i) = \sum_i \mu \{v_i\} f(v_i)$, for all $f \in A$. By choosing $f_k(x) = x_k$ (ie, the coordinate functions), we see that this implies $\sum_i \lambda_i v_i = \sum_i \mu \{v_i\} v_i$, and by uniqueness of the barycentric coordinates, we have $\mu \{v_i\} = \lambda_i = \mu_x \{v_i\}$ for all $i$, hence we have uniqueness.