[Warning: this does not intend to be a rigorous proof, expressed in technical terms, it is only an intuitive mnemonic help to better understand the power rule.]
In the end the answer to my specific question was more or less obvious. If what drives instantaneous velocity away from average velocity is “something with a number”, it is only logical that this same number provides the rule for quantifying the said separation. With a linear or “t alone” function, average and instantaneous growth rates go together. With a quadratic one like "t2", average lags behind and with a cubic one like “t3”, average lags even more. Hence it seems to follow that the degree or exponent of the function is the coefficient that must be applied onto an average rate to convert it into the instantaneous one. If we look at the reverse operation, the same logic applies: if the instantaneous rate has rocketed in proportion to the exponent, in order to return to the average rate you should “spread” the bigger rate among (i.e., divide it by) the number of causes that made it grow.
For a visual confirmation of this preliminary conclusion, I think that the analogy mentioned in the comment above is still valid: the exponent plays that role because it makes the dependent variable grow as the space occupied (area, volume and so on) by an expanding geometric shape. Let us develop this.
The function for finding displacement of an object, up to third degree rate of change (officially called “jerk”), is:
∆r = r o + v o∆t + a∆t2/2 + j∆t3/6
This is like a sum or superposition of functions, as if the object were undergoing three cumulative “states of motion”: it is static at an initial position, it is moving with an initial velocity, it is subject to a given acceleration and to a given jerk. But it is when you focus on a specific term, that you can see that its exponent acts as indicated above. For example, focusing on the 3rd degree term (“jerk”), we could represent the displacement caused by it as the volume of a 3D figure whose sides grow by 1 with every second (just read the output as “linear meters” instead of “cubic meters”).
The specific geometric figure to be taken as example depends on the “a” (acceleration) or “j” coefficients. If I am not mistaken, with an acceleration of modulus = 1, we would be in face of a right triangle with area growing as t2/2; we need acceleration = 2 to have an expanding square of growth = t2; similarly we need jerk = 6 to be in front of an expanding cube growing as per t3. But no matter the specific shape, what is important for the purpose of the analogy is that we are always in front of growth in several equal rates corresponding to the figure’s number of dimensions. Just like an expanding cube grows in 3 directions as per three equal rates, an object subject to jerk is gaining a “volume” of displacement also as per three equal rates.
Given so, how to visually guess the instantaneous growth rate? I would note in this sense that, when you make a geometric shape expand, what you do is sort of pulling it by a skin comprising each of its dimensions, nothing less, nothing more: a sort of L form (= a half-perimeter = 2t = 2 sides) in the case of 2D figures and an L of surfaces with an additional lateral wall in 3D figures (3t2), always of width = 0. Translating this image into algebra, the power rule can be formulated as follows: pull from the skin of the shape = descend 1 dimension = divide by the time elapsed and thus descend to the level of average rate = reduce the power by 1, but pull in all of the shape’s growth directions = multiply by the number of dimensions = make the average rate become instantaneous = multiply by the power.
PS: There is a good discussion here showing why 2L is the derivative of an expanding square. There they show it by pointing out that the area of the form that is added to a square when you stretch it is L∆l + L∆l (the fringes added to each side) + ∆l2 (a little square between those two fringes). With an area we are still at the level of the function (i.e. "displacement" in the motion case). To move to "average velocity" you must divide by ∆l and thus arrive at the well-known expression that appears in the difference quotient: 2L + ∆l. Final step is just stipulating that ∆l tends to 0 and thus you end up with the instantaneous rate or derivative, 2L.