In the "classical theory of topological vector spaces" the questions like this are intricated (in my opinion, this is an artifical complexity, the Nature can't be so complicated). But in the theory of stereotype spaces they become simple: for each stereotype space $X$ (including ${\mathcal S}({\mathbb R}^k)$, since it is also stereotype as a Fréchet space) the pairing $$ (x,f)\in X\times X^\star\mapsto f(x)\in{\mathbb C} $$ is a continuous bilinear form in the stereotype sense, and, as a corollary, it can be extended to a continuous linear functional on the "projective stereotype tensor product" $\circledast$ (this is an analog of $\hat{\otimes}_\pi$ in the stereotype theory) $$ \operatorname{cont}: X\circledast X^\star\to {\mathbb C}. $$ This functional is called a "contraction", you can look at the details in my paper of 2003 (page 265).
If you want to define a trace for all operators $\varphi:X\to X$ that are images of the tensors $\alpha\in X\circledast X^\star$ under the Grothendieck transformation $X\circledast X^\star\to{\mathcal L}(X)$, then your space $X$ must have the stereotype approximation property. As far as I know, nobody was interested up to now, whether the space ${\mathcal S}({\mathbb R}^k)$ has the stereotype approximation, but at first glance this is true: one can try to use the same trick as I did in my paper of 2018 for the space ${\mathcal C}(G)$ of continuous functions on a locally compact group $G$. Also Albrecht Pietsch writes in his Nuclear locally convex spaces, 10.3.2, that the Hermite polynomials form a basis in the space ${\mathcal S}'({\mathbb R})$ (which coincides with the stereotype dual space ${\mathcal S}({\mathbb R})^\star$). I think something similar must be true for each ${\mathcal S}'({\mathbb R}^k)$ (with arbitrary $k\in{\mathbb N}$), and if so, then ${\mathcal S}'({\mathbb R}^k)={\mathcal S}({\mathbb R}^k)^\star$ and ${\mathcal S}({\mathbb R}^k)$ have the stereotype approximation property.