Let $T : \mathbb R^n \rightarrow \mathbb R^n$ be a diffeomorphism.
Whenever $\vec v$ is a vector field over $\mathbb R^n$, then the push forward $T_\ast \vec v$ of that vector field along $T$ is obtained by
$T_\ast \vec v_{|T(x)} := DT_{|x} \cdot \vec v_x$.
Whenever $\omega$ is a $1$-form over $\mathbb R^n$, then the pullback $T^\ast \omega$ of that $1$-form is defined by
$T^\ast \omega_{|x} := \omega_{T(x)} \cdot DT_{|x}$.
Here, the the differential of $T$ is actually the push forward of the argument of a vector field.
I am trying to make sense of these definitions when the gradient is involved. Let $u : \mathbb R^n \rightarrow \mathbb R$ be a smooth function, and let $\nabla u$ be its gradient. Obviously,
$\nabla( u(T(x) ) = \nabla u \cdot DT_{|x}$,
so the gradient is supposed to transform like a $1$-form. This makes sense, because every differential geometer can tell that the gradient is defined as applying the musical isomorphism to the exterior derivative of a function.
Question Does the pushforward of the gradient of a function have any natural meaning? I am thinking in particular about coordinate transition between coordinate charts. Clearly, we want to transform vector fields by the push forward.