Just to flesh out some things already said in the comments.
What the sentence might be hinting at is that if you have an entangled state (for example, a two-qubit state), it might be possible to measure an outcome on one side, and obtain full information about what will be measured on the other side, even the two individual measurements are "incompatible".
More concretely, consider a simple example such as a two-qubit maximally entangled state of the form $|00\rangle+|11\rangle$. In this case, measuring the first qubit in the $Z$ basis means to perform the projective measurement with possible outcomes $|0\rangle,|1\rangle$. If you find $|0\rangle$, then the second qubit collapses to $|0\rangle$, and same for $|1\rangle$.
This means that if you also measure the second qubit in the $Z$, you get a deterministic outcome (always $0$ if the first measurement gave $0$, and same for $1$).
However, if you measure instead the second qubit in the $X$ basis, i.e. perform the projective measurement with outcomes $|\pm\rangle$, then regardless of what you found in the first measurement, you have total uncertainty about the second measurement outcome.
But the above has nothing to do with $Z$,$X$ being incompatible observables. In fact, if you repeat the same exact reasoning but with initial state $|0,+\rangle+|1,-\rangle$, you'll find the opposite result: measuring $Z$ on one side determines the result of $X$ on the other size.
There is no contradiction with uncertainty principles or the likes, because the measurements that are being performed in this scenario are really $Z\otimes I$ and $I\otimes X$, which commute.
