I am reading about the Schwartz kernel/nuclear theorem which states for any bilinear form $B$ on $\mathcal{S}(\mathbb{R}^n) \times \mathcal{S}(\mathbb{R}^n)$ (where $\mathcal{S}(\mathbb{R}^n)$ is Schwartz space), and thus an element of $\mathcal{S}'(\mathbb{R}^n)\otimes \mathcal{S}'(\mathbb{R}^n)$, there is a a continuous linear operator $K: \mathcal{S}(\mathbb{R}^n) \rightarrow \mathcal{S}'(\mathbb{R}^n)$ so that $$B(\phi, \psi) = \langle K\phi, \psi\rangle, \quad\forall \phi, \psi \in \mathcal{S}(\mathbb{R}^n)$$ where $\langle \cdot, \cdot \rangle$ is the usual pairing between a Schwartz function and its dual.
At first glance this seems very similar to the musical isomorphisms in differential geometry, or even more generally in linear algebra which states for an inner product space $V$ (of finite dimension) there is an isomorphism between $V$ and its dual $V'$ by fixing one of its arguments: $$ V \ni v \mapsto \langle v, \cdot \rangle \in V'. \tag{1}$$
Based on my readings the Schwartz kernel theorem is a big deal. Part of why this is has already been answered in this question, but I still don't see why it's so important in view of (1). Is the big result that the isomorphism (1) can be modified so that it also applies to Schwartz space (which isn't automatic since it is infinite dimensional)?