Alice receives a quantum state $|\psi\rangle$, which is an element of some basis $\mathcal{B}$, though she does not know what $\mathcal B$ is. She then teleports this to Bob, *who is told by someone else what $\mathcal B$ is*. Furthermore, it seems that specifically either $\mathcal{B} = \{ \lvert 0 \rangle, \lvert 1 \rangle \}$ or $\mathcal{B} = \{ \lvert + \rangle, \lvert - \rangle \}$.

The last detail is important here.
It means that the basis is either the eigenbasis of the $Z$ operator, or the eigenbasis of the $X$ operator — which means that when Bob wants to do the corrections, one of the two corrections which he will perform will in principle have no effect, while the other will simply interchange the two possible basis elements.

In a normal teleportation procedure, just after Alice's measurement, the state on Bob's side looks perfectly random to him.
If Bob didn't know whether Alice would be sending one of the states $\{ \lvert 0 \rangle, \lvert 1 \rangle \}$ or one of the states $\{ \lvert + \rangle, \lvert - \rangle \}$, there is very little that he can do with this qubit which would be of any use: he would have to wait to obtain Alice's measurement outcomes.
If he did wait for Alice's measurement outcomes, he could perform the necessary $X$ and/or $Z$ correction to obtain the state.

If someone (Alice or someone else) then came along and told him what
basis $\mathcal B$ his state was in, he could perform a
(non-demolition) measurement in the basis $\mathcal B$ to find out
which state it is without disturbing it.
But the outcome of this measurement is not affected by one of the corrections he did from Alice's measurement outcomes. If $\mathcal B = \{ \lvert 0 \rangle, \lvert 1 \rangle \}$, then the $Z$ correction from Alice's teleportation results didn't have any effect on the teleported state; and if $\mathcal B = \{ \lvert + \rangle, \lvert - \rangle \}$, then the $X$ correction from Alice's teleportation results didn't have any effect.
The effect of the other correction, on the other hand, would be just to interchange the two possible states of the $\mathcal B$ measurement, which could have been done before or after the measurement.

It follows that if someone told Bob what basis $\mathcal B$ Alice's state was going to be in, he could perform the $\mathcal B$ measurement first, and then wait for Alice's measurement results.
The measurement will disturb the state, but it will only collapse it to a state of the same basis $\mathcal B$ that Alice is sending; then Alice's teleportation results will give Bob enough information to correct the state.

Now, if Bob performs the measurement before Alice sends him the correction information, the result of his measurement will be random. But because the two bases are either the $X$ eigenbasis or the $Z$ eigenbases, the outcome of his measurement will either be the state he is meant to obtain or the orthogonal state; he will always know that the state he obtains is one of those two states, and what those two states are; and it will always be correctable using the information that Alice sends. (The same is also true if there is a third possibility, that the state is in the $Y$ eigenbasis, for essentially the same reason using the fact that $Y \propto XZ$.)

1I haven't watched the talk, but does he perhaps restrict the set of possible bases? If the basis B, when represented as an operator $\vec{n}\cdot\vec{\sigma}$ either commutes or anticommutes with both $X$ and $Z$, then you can certainly still proceed. For example, if B were a choice between $X$, $Y$ or $Z$, you'd be fine. Basically because you'd be guaranteed that the error $X^{M_1}Z^{M_2}$ maps basis states to basis states. – DaftWullie – 2019-03-27T08:56:10.787