The 15A location on the starter is where you want the battery and the alternator connected to. The other large lug is the side of the solenoid that is energized when the starter is operating, if you connected either terminal to that you would have a problem. If it was the battery the starter motor would be spinning all the time. If it was the alternator it would only be connected when the starter was active. Neither is a good situation. That said I'm pretty sure that's not your issue.
What it looks like to me:
Since your meter only rings if its showing less than .4 Ω its reasonable to assume that the resistance from the alternator terminal to the battery is a value that is higher than that. So what would happen if you have exactly .4 Ω's of resistance in your alternator's power connection? Assuming that V=IR is still valid and your system load is 25 amps when the car is running: .4Ω*25A= 10V. That wire would have a 10 volt drop across it!!!
.4Ω is way, way too much resistance. To maintain a charge on your battery your alternator will need to put out an extra 10 volts. This is a problem for a couple of reasons.
1: The alternator would need to produce 22.6 volts just to maintain a charge at a very low load (25 amps).
2: The regulator on our cars is not externally referenced. It only looks at the voltage is the output terminal. If there is extra resistance it has no way of knowing its not charging the battery.
What should the resistance be? 6 AWG wire is listed at .3951 Ω per 1000 feet @ 20°C. I'm going to guess that your alternator wire is shorter than that so it should be less.

For discussion lets say it's six feet. That should put your alternator to battery wire resistance at .0024Ω's. Your resistance reading is about 160 times too high.
I would measure the resistance with an ohm meter, something in that circuit has too much resistance. In my experience diagnosing voltage drops in these circuits is not intuitive what appears to be a small resistance number is actually very large when your dealing with highs currents. When in doubt do the calculations.
The normal drop on that wire at the alternators full output would be 150A * .0024Ω's = .36V.
A problem is that many meters don't measure low resistances very well. You can also measure voltage drops if the wire has a load across it.
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