Am I missing something in my understanding of the switching function of Q1?
No.
Am I missing something in my understanding of the resistor sizing?
Not in the value vs. supply voltage vs. power rating part.
More to the point, is this design bananas? Why is 130 ohms burning over 1W the right choice here? For the pull-up to 12V why wouldn't you use a much higher value? Or is there some other more typical way of designing a bit of circuit here to convert from TTL levels to 0/12V such that you're not burning crazy power whenever the RAM is not enabled?
No, the design isn't bananas, at least not for the mid 1970's. What you're missing is the input capacitance to the DRAM chips. I don't know what that is, but take the input capacitance to the chip, multiply it by 8 (for the 8 CE lines), add in another 50% to 150% for stray capacitance, then figure out the RC time constant -- compare that to the specified rise and fall times of the chip's CE line.
The designer chose to use a transistor driver instead of a push-pull driver for some reason. This may have been because that's what they knew, it may have been because there weren't drivers strong enough to do the job, it may have been for cost reasons (because logic was expensive back then).