$ \left( \dfrac{\partial}{\partial x} \right)_{p} $ is both an element of the tangent space $ {T_{p}}(M) $ and a linear functional on $ {C^{1}}(M) $, while $ (\mathrm{d}{x})_{p} $ is an element of the cotangent space $ {T^{*}_{p}}(M) $.
Question. How should I interpret $ \dfrac{\partial^{2}}{\partial x^{2}} $, or more generally, higher-order partial derivatives?
Let $ (M,\mathcal{A}) $ be an $ n $-dimensional $ C^{\infty} $-manifold, meaning that $ M $ is a topological space and $ \mathcal{A} $ is a set of ordered pairs $ (U,\phi) $, where $ U $ is an open subset of $ M $ and $ \phi: U \to \Bbb{R}^{n} $ is a topological embedding, such that $ \displaystyle M = \bigcup_{(U,\phi) \in \mathcal{A}} U $ and for $ (U,\phi),(V,\psi) \in \mathcal{A} $ with $ U \cap V \neq \varnothing $, the map $$ \psi \circ \phi^{-1}: \phi[U \cap V] \to \psi[U \cap V] $$ is a $ C^{\infty} $-diffeomorphism between non-empty open subsets of $ \Bbb{R}^{n} $.
Note: We call $ \mathcal{A} $ an atlas and each element of $ \mathcal{A} $ a coordinate chart.
Next, a function $ f: M \to \Bbb{R} $ is called $ C^{\infty} $ if and only if for each $ (U,\phi) \in \mathcal{A} $, $ f \circ \phi^{-1}: \phi[U] \to \Bbb{R} $ is a $ C^{\infty} $-function from an open subset of $ \Bbb{R}^{n} $ to $ \Bbb{R} $.
For a fixed $ (U,\phi) \in \mathcal{A} $, write $ \phi $ in component form: $ \phi = (x_{1},\ldots,x_{n}) $. Hence, $ x_{i} $ is a map from $ U $ to $ \Bbb{R} $ for each $ i \in [n] $. For each $ C^{\infty} $-function $ f: M \to \Bbb{R} $ and each multi-index $ \mu $, we define a function $ \dfrac{\partial^{\mu} f}{\partial x^{\mu}}: U \to \Bbb{R} $ by $$ \forall p \in U: \quad \frac{\partial^{\mu} f}{\partial x^{\mu}} (p) \stackrel{\text{df}}{=} \left[ \mathbf{D}^{\mu} \! \left( f \circ \phi^{-1} \right) \right] \! (\phi(p)). $$ Hence, the ‘higher derivatives’ of $ f $ depend on the coordinate chart that you are using.