The lesson here is that graphical intuition isn't always the best choice. For example, you can say that the intuition for derivatives is that they're slopes, and for integrals is that they're areas. But why would "slopes" be that useful in physics? I mean, besides inclined planes, you don't see that many literal slopes in a physics class. And why "areas"? We live in 3D space, so shouldn't volumes be more important?
In case that all sounds dumb, the point is that sometimes, you can actually make something less physically intuitive by explaining it visually, because usually the visual explanation is completely devoid of the dynamic context that would be present in a real physics problem.
One of the main reasons derivatives and integrals appear so often in introductory physics is that they're with respect to time, so the derivative means "a rate of change" and the integral means "an accumulation over time". This is a distinct intuition from the geometric one. The point of the geometric intuition is to help you see what the derivative and integral is given a graph, but it doesn't really help you interpret what it physically does.
Similarly, there is a complicated geometric intuition for the convolution, which can hypothetically help you eyeball what the convolution of two graphs of functions would look like. But in this case the "dynamic" intuition is much simpler.
One piece of intuition
Convolutions occur whenever you have a two-stage process where the stages combine linearly and independently.
Suppose that I kick an initially still mass on a spring at time $t = 0$, and the subsequent trajectory of the spring is $x(t)$. If I apply the same kick at time $t = 1$, then by time translational invariance, the subsequent trajectory is $x(t-1)$. Now suppose that I kick both at $t = 0$ and $t = 1$, with strengths $f(0)$ and $f(1)$. Then by linearity, the subsequent trajectory is
$$f(0) x(t) + f(1) x(t-1).$$
This is like the "reversing and shifting" structure of a convolution. So more generally, if I apply a continuous force $f(t)$, then the trajectory is
$$\int dt' \, f(t') x(t-t')$$
which is precisely a convolution. This is where a large fraction of the convolutions in physics and electrical engineering come from. (A large fraction of the remainder come from the fact that the Fourier transform of the convolution is the product of the Fourier transforms, and products are simple and ubiquitous.)