Can we actually walk along the gradient of a scalar to climb the hill faster?

Solution 1:

Your function acts from $\mathbb{R}^2$ to $\mathbb{R}$ (the height of a given point). Through the point you have chosen, you can draw a circle on which the function is constant (assuming that the hill has the appropriate shape). The vector (in the plane $\mathbb{R}^2$) pointing to the center of the hill is really perpendicular to this circle at this point.

That is, your mistake is that you misunderstood what "constant height surface" means. We are talking about a surface (dimension $n-1$) in the domain of the function, and not anywhere in its graph.

Solution 2:

This image colors vectors to represent their length to cut back on clutter (warmer colors represent longer vectors). Those vectors constitute the gradient, and they point horizontally, whereas the surface is $3$-dimensional. So what is the relation between the surface and its gradient?

For an intuitive idea, imagine a video game character not particularly susceptible to gravity, such as an ant, standing on either the inside or the outside of the surface. The vector directly beneath your character tells you how to hold the control stick in order to move him in the direction of steepest ascent (assuming the game's camera cooperates). The length (here color) of that vector tells you how steep it will be. This should give an intuition on how movement along a $3$-dimensional function can be controlled using only $2$ dimensions.