Here's an attempted expansion.
If you have an equation of the form dy/dx = f(x) and you want "solve" it, what you are typically looking for is to write y = g(x), right? In other words, the solution to this differential equation is some curve in the x-y plane. This applies more generally, e.g. to situations where you end up with an "implicit" solution like h(y) = g(x): you still get an equation relating x and y which can then be represented as some set of points in the x-y plane (the ones that satisfy that equation).
Now say f(x) happens to have the form a(x,y)/b(x,y). You can consider the system of two differential equations: dy/dt = a(x,y), dx/dt = b(x,y). Solving this system gives x and y as functions of t. Picking any particular value of t gives values of x and y, which gives you a point in the x-y plane. The first key point is that the set of points produced by this procedure as you plug in all possible values of t is exactly the set of points for which the h(y) = g(x) equation above holds. In other words, the solution to the two-equation system encapsulates all the information about the solution to the original equation.
The second key point is that the solution to the two-equation system has _more_ information than the solution to the original equation. In particular, it has the actual values of dx/dt and dy/dt for every given value of t, which don't correspond to anything in our original problem. Their _ratio_ does correspond to something in our original problem: the slope of the tangent line to the solution curve (dy/dx). But the exact values themselves are somewhat arbitrary, as long as their ratio is correct. Put another way, our original problem's solution is a curve in the x-y plane, while the solution of our two-equation system is a curve together with a description for how fast to move along it as t changes. That's the "velocity" bit in Rota's article.
OK, but if how fast we move along the curve doesn't really matter, maybe we can choose to move along it in a nice way that makes it particularly simple to figure out what the shape of the curve is. Our only constraint is that at any given point along the curve the ratio of dx/dt and dy/dt is fixed, because in our original problem we have a fixed dy/dx if we're given values of x and y. So if, at every point (x,y) we multiply dx/dt and dy/dt by the same number (which can depend on x and y) then we get a system of two equations that has different solutions for x and y as functions of t, but the graph of the resulting thing in the x-y plane still looks the same. That's the integrating factor bit; we just formalize it by saying that we multiply both dx/dt and dy/dt by the same function q(x,t), which is exactly what it means to multiply them both at every point by some number that might depend on that point.
The hard part, of course, is choosing a q(x,y) that makes things work out nicely and makes it easy to solve our two equations to get x(t) and y(t).
Here's a concrete example that might help:
Say dy/dx = x/y. We rewrite this in the form dy/dt = x, dx/dt = y. This isn't terribly convenient to solve, so we multiply by q(x,y) = 1/(2xy) to get a new system: dy/dt = 1/(2y), dx/dt = 1/(2x). At this point, maybe you just look at it and go, ah, y = sqrt(t + C1), x = sqrt(t + C2), or maybe you figure out some other way to get there. In any case, now you see that t + C1 = y^2, t + C2 = x^2, so x^2 - y^2 = C for some constant (C2-C1, but both are arbitrary, so this is just some single arbitrary constant). And that's your (implicit) solution for the original differential equation: a hyperbola, or more precisely a family of hyperbolas each of which satisfies the equation.
To illustrate the point about velocities, let's just consider C = 1, so x^2 - y^2 = 1. The point (sqrt(2), 1) lies on this curve. At this point, dy/dx = x/y = sqrt(2). On our original formulation of the parametric system, dy/dt = sqrt(2), dx/dt = 1 at this point. In our reformulation with the integrating factor, dy/dt = 1/2 and dx/dt = 1/(2*sqrt(2)). So the two formulations have us moving along the hyperbola at different speeds at this point as t changes, but they're moving along the same hyperbola.
Does that help at all?