1

If you fix $y$, the function $x \mapsto \mathrm{grad}_y f(x,y)$ is a function mapping $M \to T_y N$; note that the codomain is a single, fixed vector space. So the differential in $x$ is a mapping $TM \to T(T_yN)$, but as $T_yN$ is a linear space its tangent space is canonically isomorphic to itself.

2

They should always be adjoints. You have, extending $\eta$ to a constant map $M \to T_yN$, $$ \langle \eta, \mathrm{grad}_{xy}^2(\xi)\rangle_y = \xi( \langle \eta, \mathrm{grad}_y f(x,y) \rangle_y) = \xi(\eta(f)) $$ So: given a vector $\eta\in T_yN$ and a vector $\xi$ in $T_xM$, extend them to a vector field $\eta$ on $N$ and a vector field $\xi$ on $M$, and then extend them trivially to vector fields on $\tilde{\eta} = (0,\eta)$$\tilde{\eta}(x,y) = (0,\eta(y))$ and $\tilde{\xi} = (\xi,0)$$\tilde{\xi}(x,y) = (\xi(x),0)$ on $M\times N$. Then our computation above shows that the difference between the two expression you are asking about is exactly $[\tilde{\xi},\tilde{\eta}]f$, but it is easy to check that this Lie Bracket vanishes.