Well, relative motions are more intuitive to me - they make sense, and I can use calculations for them.

In the first example, you presented 101 speed - this means only 1 speed relative to the planet, and that's all that's getting redirected (in the planet's frame of reference your enter and exit velocity should be the same, since that's how orbits work). The number is just too small, but your velocity would be planet velocity + 1 on a different vector, which will be less than 101 total.

If we estimate the angle on the picture is about 60 degrees from the velocity vector, and the speed to be 100+v1, the speed from the planet's frame of reference is v1 - so, the exit velocity will have components of (100+v1cos(60°)) and (v1sin(60°)), so the final speed relative to the sun should be

sqrt((100+x*cos(60°))^2+(x*sin(60°))^2)

Wolfram alpha suggests this simplifies to sqrt(x(x+100)+10000), and comparing the equation by appending <x+100 gives the solution of x>0

This means, if my math is correct, with an entry angle of 0° and exit angle of 60°, you always lose speed.

I could try replacing the angle with a variable and setting a constraint of x>0 and see if the free version of wolfram alpha would spit out something, but just replacing the 60 with y is spitting out some convincing solutions, since in those x is never greater than 0.

PS: Thanks for taking the time to explain, the fact that you went out of your way to draw the diagrams is not lost on me. I unfortunately still don't see it, but I really do appreciate the effort!