There is much more information provided here which may help.
Let me start by clarifying that the aim is not to detect and correct logical errors, but to detect and correct physical errors, i.e. those that occur on single physical qubits rather than the encoded qubit.
At the macro level, the answer is that if you have an $X$ error (say), you use a controlled-not to propagate it to another qubit where it appears as an $X$ error. Measuring in the $X$ basis cannot see that error. For example, if a single qubit state was in $|+\rangle$, an $X$ error would not change it, and you would be unable to detect the error. On the other hand, if the qubit was initially in $|0\rangle$ then the possibility of a flip would put it in either $|0\rangle$ or $|1\rangle$, which can be detected with a $Z$ measurement. (Put another way, $X$ errors commute with $X$ measurements, so they do not affect each other.)
More specifically, let's start by thinking about what happens if there is no error. In that case, $|+\rangle$ is the eigenstate of controlled-not, so it never changes. This is what initially threw me trying to think about a $Z$ measurement on this. But of course, it's not logical $Z$ measurement, but physical. So, basically, you'll get a single answer corresponding to one of the basis states used in either to 0 or 1 logical state, with no control over which. Let's call that answer $x$. What we do know is that $H\cdot x=0$, where $H$ is the parity-check matrix of the code (I'm being a bit loose about which of the two it is).
Now, what happens if there was a single physical $X$ error somewhere on the logical qubit. Controlled-nots (and we should think about the individual physical ones here, not the logical ones) propagate $X$ rotations from control to target. So, instead of getting an answer $x$, we'd get an answer $x\oplus e$ where $e$ is a vector representing the single qubit that had the error. So, now, if you apply the parity-check matrix, you get
H\cdot(x\oplus e)=(H\cdot x)\oplus(H\cdot e)=H\cdot e.
The whole point of the parity-check matrices is that they can let you identify any single-qubit $e$. Hence, you know what correction to provide.