@DavidHammen and I agree: that delta-V plot is at best misleading, and at worst — well, Dave is a smart fellow, and he wisely didn't want to use the word here on SESE. It turns out the delta-Vs for all these transfers depend tremendously on how you do them.
An example: as you point out, the plot says the delta-V to go from "Earth C3=0" to "Mars transfer" is 0.6 km/s. I won't go into the differences among the various combinations of departing Earth at perihelion or aphelion or somewhere between, and arriving at Mars with Mars at perihelion or aphelion or somewhere between; for now I'll just assume those orbits are circular, but I'll assume that we get to Mars at a heliocentric distance of 208 million km, as you did; the precise distance won't make a qualitative difference in the result. Let's look at two ways of doing the transfer orbit injection.
The first is to actually get out to where C3=0 takes you, escaped from Earth. You're orbiting the sun now at the same speed as Earth, and with the circular-orbit approximation that's 29.78 km/s. You need to burn to the transfer orbit, which has a perihelion velocity of 32.12 (plus a bit) km/s. So you need to speed up by 2.34 km/s, and of course that's the delta-V to get onto that transfer orbit.
The second is to assume you're in a C3=0 orbit but you're at perigee (I'll assume 200 km altitude), and you do your TMI (trans-Mars injection) burn there. When you escape from Earth you'll need your V-infinity to be the same 2.34 km/s we saw above. But now you're down deep in Earth's gravity well, so the Oberth effect comes to your aid. At 200km altitude, a C3=0 orbit has a velocity of 11.01 km/s. An Earth escape orbit with a V-infinity of 2.34 km/s has a perigee velocity (assuming the same 200 km altitude) of 11.25 (plus a bit) km/s. So getting assistance from Hermann Oberth, you could actually go from C3=0 to TMI for only 0.24 (plus a bit) km/s! That's about one tenth of the free-space delta-V!
That Wikipedia chart does not consider this at all! And that is why Dave and I take exception to it. Delta-V is not linear!! Any chart that implies it is, is ... well ... I won't use the word either, Dave.
Exercise for the student: assuming an impulsive delta-V precisely aligned with the velocity vector, and assuming the orbits as I described above, at what orbit altitude (or geocentric radius, if you prefer) would you have to perform the TMI burn to make the needed burn magnitude 0.600 km/s??
