Hessian (as the second covariant derivative) of a smooth function

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Consider the vector bundle $$\pi\colon \underline{\mathbb R}\to M$$ where $\underline{\mathbb R}$ is the trivial line bundle and $M$ is some smooth manifold. Clearly, the sections of $\underline{\mathbb R}$ are the $C^\infty(M)$-functions $f\colon M\to \mathbb R$.

Now let $\nabla^{\underline{\mathbb R}}$ be a connection on $\underline{\mathbb R}$ and let $\nabla^{TM}$ be a connection on the tangent bundle $TM$. We write $\nabla^{\underline{R}}f = df$.

Then we defined the Hessian as the second covariant derivative of $f$ via

$$\nabla^2_{X,Y}f = (\nabla_Xdf)(Y) = XYf - (\nabla Y)f\quad (*)$$


Question 1:

Am I correct in assuming that the above expression $(*)$ is supposed to be

$$\nabla^2_{X,Y}f = (\nabla_Xdf)(Y) = XYf - (\nabla_\color\red{X} Y)f$$ where $X,Y$ are vector fields on $M$?

Question 2:

If I would like to obtain $(*)$, I suppose it follows from the product rule in the sense that

\begin{align} \nabla_X(df\otimes Y) &= (\nabla_Xdf)\otimes Y + df\otimes\nabla_XY \\\iff \underbrace{(\nabla_X)(Yf)}_{XYf} &= \underbrace{(\nabla_Xdf)(Y)}_{\nabla^2_{X,Y}f} + (\nabla_XY)f \end{align} and hence

$$\\ \nabla^2_{X,Y}f = XYf - (\nabla_XY)f$$

Since I am always really confused with notations and computations in differential geometry, I'd really appreciate if someone could provide some feedback whether my computations (and assumptions) are correct.

Thanks for any help!