I think the argument here is twofold: 1) If you have some first-preimage attack against a hash, that attack can probably be "continued" such that it produces m

I think the argument here is twofold: 1) If you have some first-preimage attack against a hash, that attack can probably be "continued" such that it produces multiple preimages. 2) Even if your attack can only ever produce one preimage, since there's an infinite number of documents that can produce the same hash, it's vanishingly unlikely that your attack will produce the exact preimage that you already have (note: this is assuming a cryptographically secure hash, where all outputs are equally likely). Therefore, even if you can only get one preimage, it's still almost certainly a second preimage.