Let $V \subset H$ be a continuous embedding of Hilbert spaces. Let $I\colon V \to H$ be the inclusion map. Let $R\colon H \to H^*$ be the Riesz map. Then we have for $h \in H$ and all $v \in V$ $$\langle I^*Rf, v \rangle_{V^*,V} = (f,v)_H$$ where $I^*\colon H^* \to V^*$ is the adjoint map of $I$.
Now let us identify $H$ with its dual and make $V \subset H \subset V^*$ a Gelfand triple. In this case, we would write for any $h \in H$ and $w \in V$ the formula $$\langle h, w \rangle_{V^*,V} = (h,w)_H.$$
So is the whole point of the Gelfand triple that it means we don't need to write $I^*R$ in the first displayed equation? It is just a notation?
In Quantum Mechanics Beyond Hilbert Space, Jean-Pierre Antoine explained that the rigged Hilbert space formalism gives quantum theorists access to generalized eigenvectors of (extensions of) linear operators that have continuous spectrum. These generalized eigenvectors can be used to represent a variety of operations in a more familiar setting and notation.
In particular, if the rigged Hilbert space is $\Phi\subset\mathcal{H}\subset\Phi^{\times}$, then each self-adjoint operator on the Hilbert space $\mathcal{H}$ has an extension that has a complete orthonormal set of generalized eigenvectors in $\Phi^{\times}$, and inner products of vectors in $\Phi$ can be expressed as integrals with respect to a (conventional?) measure on $\mathbb{R}$.
I should note that Antoine assumed that $\Phi^{\times}$ consists of antilinear functionals on $\Phi$ that are continous with respect to a topology with respect to which $\Phi$ is complete.