Let's say that MiningPoolA controls 51% of hashpower, and MiningPoolB controls 49%. MiningPoolA is refusing to mine transactions from/to some wallet W.
Time T1:
MiningPoolA: OldChain -- Block1(no W)
MiningPoolB: OldChain -- Block1'(W receives 1BTC) still in progress
Any client will accept the chain with Block1 (no transactions from/to W)
Time T2 - if MiningPoolB tries to compete
MiningPoolA: OldChain -- Block1(no W) -- Block2 (no W)
MiningPoolB: OldChain -- Block1'(W receives 1BTC) -- Block2' (parent=Block1') still in progress
Any client will accept the chain with Block1 (no transactions from/to W), and MiningPoolB will never be able to catch up
Time T2 - if MiningPoolB decides to accept MiningPoolA's chain, but add the transaction in the second block:
MiningPoolA: OldChain -- Block1(no W) -- Block2 (no W)
MiningPoolB: OldChain -- Block1(no W) -- Block2' (parent=Block1, adds transaction W receives 1BTC) still in progress
Any client will accept the chain with Block1 (no transactions from/to W), and MiningPoolB will never be able to catch up
If a mining pool had 51+% of hashpower, they would always be mining the longest chain, no one would be able to compete with them and publish another block (in principle, at least; in practice, since mining is not entirely deterministic, someone else will occasionally win the lottery and propose a new block faster).The 51% chain is mined with 51% of the total mining power, while the 49% chain will contain 100% of the mining power (as the 49% miners are more than happy to build off the 51% chain, while the 51% miners will only building off the chain without the censored transactions).
They are correct in asserting that a 51% attack to censor transaction is largely just a nuisance to the users and that all transactions should eventually end up on the longest chain (the 49%).
Each miner takes a bunch of transactions and chooses a previous block B1 to build on, then starts hashing. If some other minerB advertises a new block B2 based on B1 that includes different transactions before minerA, then minerA can throw away all the work it did, and start from scratch on B3 based on B2. But minerB will probably already be working on its own B3 - and has every chance to win again.