Can you give a simple, intuitive explanation (imagine you're talking to a schoolkid) of why mathematically speaking momentum is covector? And why you should not associate mass (scalar) times velocity (vector) and momentum (covector) in your head?
There were already answers to this question, but they all rely on rather abstract definition of a Lagrangian (difficult to explain to a schoolkid). Is there some motivation in the first place?
Suppose you exert a force ${\bf F}$ on a particle of mass $m$ in order to move it along a path $\gamma$. The work required to do so is $$W = \int_\gamma {\bf F} \cdot d{\bf x} = \int_\gamma \frac{d{\bf p}}{dt} \cdot d{\bf x} = \frac{d}{dt} \int_\gamma {\bf p} \cdot d{\bf x}$$ (since the path $\gamma$ and therefore $d{\bf x}$ does not change over time).
Now the work done is a scalar; clearly it doesn't depend on your choice of spatial coordinates. You don't feel twice as tired after exerting the same force on the same physical system described using different coordinates. (More quantitatively, if you use a motor to push the mass, then you can measure the work done by the motor by seeing how much energy has been depleted from its battery. Clearly the battery doesn't "know" what coordinate system you used.) And the time derivative and the line integral also clearly do not depend on the choice of spatial coordinates. Therefore, the dot product ${\bf p} \cdot d{\bf x}$ must be coordinate-independent as well. So if you change coordinates in such a way that the numerical values of the components of $d{\bf x}$ double, then the numerical values of the components of the momentum ${\bf p}$ must halve in order to compensate. So the position and momentum transform oppositely under coordinate transformations (as a contraviant vector and covector, respectively).
If you don't like the integral or the time derivative in that explanation, you can alternatively note that the instantaneous power $P = {\bf F} \cdot {\bf v}$ that you apply to the mass is a scalar for the same reason, so force (the time derivative of momentum) and velocity (the time derivative of position) must transform oppositely under coordinate transformations.
Since the initial posting of my question, a reformulation has appeared with the perfect answer. Although this answer does not address the issue of explainability to a school kid, it nevertheless should help you get the mathematical gist of the concept. Hopefully, a (very) motivated school kid can understand it :) Therefore, the question is closed now.