You’re right that the analogy between elementary-school-multiplication and convolution only works for multiplication without carrying. That still leaves a large amount of possible multiplication instances, for the term “cherry picking” to apply here.
You’re wrong to suggest that elementary-school-multiplication and (discrete) convolution are so dissimilar that the analogy is so unhelpful that it only adds to everyone’s confusion. Maybe to yours, but not to everyone’s.
If I understand your objection correctly (other than the already conceded fact that the analogy does not work in case of multiplication with carrying), you already have internalized that convolution is an operator on 2 functions, resulting in a function. And since you have already internalized that, the analogy with any operator on two natural numbers resulting in a natural number confuses you — although I assume you can see the similarity in the sliding operation.
For people who have not internalized what a convolution is, the visualisationg of the sliding operation, and the analogy with elementary-school-multiplication is the helpful bit.
To clarify why the operations are really similar: consider a natural number as a function that takes a natural number as argument, and returns the digit (between 0 and 9 inclusive) for that decimal position.
So f = 2 031 is considered to be: n → f(n) = if n = 3 then return 2 else if n = 1 then return 3 else if n = 0 then return 1 else return 0 (end if).
And g = 320 is considered to be: n → g(n) = if n = 2 then return 3 else if n = 1 then return 2 else return 0 (end if).
What is then the discrete convolution f ∗ g?
It’s f ∗ g = n → (f ∗ g)(n) = f(3) × g(n – 3) + f(1) × g(n – 1) + f(0) × g(n)
= n → if n = 5 then return 6 else if n = 4 then return 4 else if n = 3 or n = 2 then return 9 else n = 1 then return 2 else return 0 (end if).
This is how we consider 649 920 = 2 031 × 320.