Let $\mathcal M$ be a Riemannian manifold and $f : \mathcal M \to \mathbb R$ be smooth. Take a smooth curve $\boldsymbol \gamma : I \to \mathcal M$, where $I$ is an open interval of the real line. I'm interested in the composition $f \circ \boldsymbol \gamma : I \to \mathbb R$.
The first derivative can be computed via the chain rule as
\begin{equation} \frac{\text{d}}{\text{d}t} (f \circ \boldsymbol \gamma) = \frac{\partial f}{\partial \boldsymbol \gamma} \dot{\boldsymbol \gamma}. \end{equation}
If I understand correctly, the first factor is the Jacobian of $f$, which in this case is a linear functional, mapping tangent vectors to scalars. Thus, the whole expression is the directional derivative of $f$ along $\dot {\boldsymbol \gamma}$. All good.
Differentiating once again, using the chain and product rule, I get
\begin{equation} \frac{\text{d}^2}{\text{d}t^2} (f \circ \boldsymbol \gamma) = \frac{\partial^2 f}{\partial \boldsymbol \gamma^2} \dot{\boldsymbol \gamma} \dot{\boldsymbol \gamma} + \frac{\partial f}{\partial \boldsymbol \gamma} \ddot{\boldsymbol \gamma}. \end{equation}
What kind of object is $\frac{\partial^2 f}{\partial \boldsymbol \gamma^2}$? For the whole expression to be a scalar, I reckon we need $\frac{\partial^2 f}{\partial \boldsymbol \gamma^2} \dot{\boldsymbol \gamma}$ to be a linear functional. But since $\dot{\boldsymbol \gamma}$ is just a vector, how would $\frac{\partial^2 f}{\partial \boldsymbol \gamma^2}$ transform a vector into a linear functional?
Also, the second term is the geodesic curvature, and vansishes if $\boldsymbol \gamma$ is a geodesic, right?