Some mathematical statements then an intuition statement:
Mathematical
We often use parameterizations like this to mathematically represent a continuous motion from one point to the next, for instance when discussing convex spaces, in which we say a space is convex if all points between any two given points are also in the set (in the language of the wiki page, convex if the vector $\lambda u_i+(1-\lambda) u_j$ is also in the space, for any vectors $u_i$ and $u_j$ in the space; satisfied by a sphere, but not by a donut).
Similarly, the parameterization here is simply a way to represent traveling through a space, in this case a space of possible states. The "nonphysical" parameter $\lambda$ isn't introducing new physics any more than my above analogy means the space of a sphere is in any way physically changed by our mathematical wandering through the space of a sphere. The reason this works is that you're wandering through a continuously changing energy landscape, which brings me to...
Intuition
For state variables like energy, as long as you wander from one state to another, keeping track of your progress as you go (more on that below), you can find your new energy from your old, just like with the traveling analogy already discussed.
The nuance is that if you, say, travel from one city to another city directly North, you might take a path that veers a bit east before coming back west to arrive at its destination. You might say, doesn't that increase my total distance traveled? Yes, but any travel east cancels any travel west. Similarly, going through this space of states with varying energies, any increases along the path cancel any decreases along the path, and you can arrive at the total energy difference between your initial and final states. Now you can see why we need this path to be differentiable with respect to $\lambda$: we can't allow discontinuous jumps.
So in conclusion, we don't really care about the nature of the path, since any funny deviations from point A to point B will cancel. $\lambda$ is just the mathematical parameter that allows us to continuously follow our path and ensure we end where we want to go.
When the particle q is approaching the other particle, the distance r1 is decreasing. The kinetic energy connected to the radious depends on the derivative of r1. At the minimum distance r1 is no longer decreasing and assumes his lower value. After that point r1 start increasing when the particle moves away. So at the beginning the derivative of r1 is negative, at the end is positive. If we assume that the derivative of r1 is a continuos function, it must be 0 when r1 is the minimum.
Best Answer
If you have a force [in 1D - the generalization in 3D is straightforward but you need vector calculus] $F(x)$ which is conservative then the potential energy is defined as
$$V(x) = - \int _{x_0}^{x} F(x')dx'$$ where $x_0$ is an arbitrary starting point.
This means, that, by the fundamental theorem of calculus, you can "eliminate" the integral by taking the derivative of the potential
$${dV \over dx} = -F(x)$$
so the condition that the derivative of the energy is $0$ means
$${dV\over dx} = 0 = F(x)$$ meaning that the total force acting on your mass is $F(x)=0$ i.e. the mass is at equilibrium (no forces are acting).
If you then wonder whether the equilibrium is stable (i.e., if you move a bit you come back where you started) then you need to look at the second derivative of the potential energy. You can check more details here.