I know what the second differential of $f : \Bbb R^n \to \Bbb R$ means. Nevertheless, when working with abstract manifolds and in the absence of a connection, one cannot come up with a 2-covariant tensor that could reasonably be called $\Bbb d ^2 f$. This is why one gets around this by viewing $\Bbb d ^2 f$ as the differential $\Bbb d (\Bbb d f) : T(TM) \to \Bbb R$ of $\Bbb d f : TM \to \Bbb R$.
In a bit of spare time I decided to try this on $\Bbb R^n$ and see what it gives. Unfortunately, the result lost me, because I do not know how to interpret it. Here it is. If $(\Bbb d _x f) (v) = \sum \limits _i (\partial _{x_i} f) (x) v_i$, then
$$\begin{align} \Bbb d _{(x,v)} (\Bbb d f) &= \sum _j \partial _{x_j} \left( \sum \limits _i (\partial _{x_i} f) (x) v_i \right) \Bbb d x_j + \sum _j \partial _{v_j} \left( \sum \limits _i (\partial _{x_i} f) (x) v_i \right) \Bbb d v_j \\ &= \sum _{j,i} (\partial _{x_j} \partial _{x_i} f) (x) v_i \ \Bbb d x_j + \sum _{j,i} (\partial _{x_i} f) (x) \delta _{ij} \ \Bbb d v_j \\ &= \sum _{j,i} (\partial _{x_j} \partial _{x_i} f) (x) v_i \ \Bbb d x_j + \sum _i (\partial _{x_i} f) (x) \ \Bbb d v_i . \end{align}$$
There are two things bothering me here:
the presence of those parasitic $v_i$ in the first sum; they should somehow play the role of $\Bbb d x_i$;
the second term, containing derivatives of order 1.
I guess that the first question is answered by noting that $\Bbb d (\Bbb d f)$ and $\Bbb d ^2 f$ are not really identical, that in fact they just correspond to each other by the isomorpshim that takes the map $(X, V) \to \Bbb d _{(x,v)} (\Bbb d f) (X, V)$ to... to what? I already have 4 arguments $x,v,X,V$, while $\Bbb d^2 f$ should have only 3. Equally important, how to eliminate the second sum?