I am recording some observations that I made while trying to understand Professor Rieffel’s definition of $ \partial_{k} $, which I managed to do in the end. :)
Let
$ G $ be a finite-dimensional Lie group,
$ {\frak{g}} $ the Lie algebra of $ G $,
$ A $ a Fréchet space (over $ \mathbb{C} $) and
$ \alpha $ a strongly continuous group action of $ G $ on $ A $.
Define $$ A^{\infty} \stackrel{\text{def}}{=} \lbrace a \in A ~|~ \pi(\bullet,a): G \to A \text{ is a smooth function} \rbrace, $$ which we call the space of smooth vectors for the action $ \alpha $ of $ G $ on $ A $. It is a linear subspace of $ A $, and by considering the Gårding space for $ \alpha $, it can be shown to be dense in $ A $.
For each $ v \in {\frak{g}} $, define a linear subspace $ {\frak{D}}(v) $ of $ A $ by
$$
{\frak{D}}(v) \stackrel{\text{def}}{=} \left\lbrace a \in A ~ \Bigg| ~
\lim_{t \to 0} \frac{\alpha(\exp(tv),a) - a}{t} \text{ exists} \right\rbrace.
$$
For each $ v \in {\frak{g}} $, we have $ A^{\infty} \subseteq {\frak{D}}(v) $. This then allows us to define a mapping $ \pi: {\frak{g}} \to \mathcal{L}(A^{\infty},A) $ by
\begin{align} \forall v \in {\frak{g}}, ~ \forall a \in A^{\infty}: \quad [\pi(v)](a) &\stackrel{\text{def}}{=} \frac{d}{dt} \Bigg|_{t=0} \alpha(\exp(tv),a) \\ &= \lim_{t \to 0} \frac{\alpha(\exp(tv),a) - a}{t}. \end{align}\begin{align}
\forall v \in {\frak{g}}, ~ \forall a \in A^{\infty}: \quad
[\pi(v)](a)
& = \frac{d}{dt} \Bigg|_{t=0} \alpha(\exp(tv),a) \\
& = \lim_{t \to 0} \frac{\alpha(\exp(tv),a) - a}{t}.
\end{align}
In fact, we have $ \pi: {\frak{g}} \to \mathcal{L}(A^{\infty},A^{\infty}) = \text{End}(A^{\infty}) $, and with a little more work, one can show that this is a Lie-algebra homomorphism, i.e.,
$$
\forall v,w \in {\frak{g}}: \quad \pi([v,w]_{\frak{g}}) = [\pi(v),\pi(w)]_{\text{End}(A^{\infty})}.
$$
In the case where $ G = \mathbb{R}^{d} = {\frak{g}} $, we have (after fixing an ordered basis $ (X_{1},\ldots,X_{d}) $ of $ \mathbb{R}^{d} $)
\begin{align} \forall k \in \lbrace 1,\ldots,d \rbrace, ~ \forall a \in A^{\infty}: \quad [\pi(X_{k})](a) &= \frac{d}{dt} \Bigg|_{t=0} \alpha(\exp(t X_{k}),a) \\ &= \lim_{t \to 0} \frac{\alpha(\exp(t X_{k}),a) - \alpha(\exp(0_{\mathbb{R}^{d}}),a)}{t} \\ &= \lim_{t \to 0} \frac{\alpha(t X_{k},a) - \alpha(0_{\mathbb{R}^{d}},a)}{t} \\ &= \lim_{t \to 0} \frac{{f_{a}}(t X_{k}) - {f_{a}}(0_{\mathbb{R}^{d}})}{t} \\ &= {D_{X_{k}} f_{a}}(0_{\mathbb{R}^{d}}). \end{align}\begin{align}
\forall k \in \lbrace 1,\ldots,d \rbrace, ~ \forall a \in A^{\infty}: \quad
[\pi(X_{k})](a)
& = \frac{d}{dt} \Bigg|_{t=0} \alpha(\exp(t X_{k}),a) \\
& = \lim_{t \to 0} \frac{\alpha(\exp(t X_{k}),a) - \alpha(\exp(0_{\mathbb{R}^{d}}),a)}{t} \\
& = \lim_{t \to 0} \frac{\alpha(t X_{k},a) - \alpha(0_{\mathbb{R}^{d}},a)}{t} \\
& = \lim_{t \to 0} \frac{{f_{a}}(t X_{k}) - {f_{a}}(0_{\mathbb{R}^{d}})}{t} \\
& = {D_{X_{k}} f_{a}}(0_{\mathbb{R}^{d}}).
\end{align}
Then by $ \partial_{k}: A^{\infty} \to A^{\infty} $, Professor Rieffel simply means $ \pi(X_{k}) $.
