Verifying Shor's quantum error correction code
Since I don’t understand what you don’t understand, I will try to give a quite formal description, hoping it will help.

$\newcommand{ket}[1]{\left|#1\right&gt;}$
Let’s suppose the error is the unitary $U$ applied on the $k$th qubit.
We have
$$\begin{align}
U&amp;=aI+bX+cY+dZ\\&amp;=aI+bX+icXZ+dZ,
\end{align}$$
with $a,b,c,d$ being real numbers, $I$ is the identity, $X, Y, Z$ the Pauli operators. $X$ is a bit-flip, and $Z$ is a phase-flip.
$Y=iXZ$ is a combination of phase and bit-flip.

Let’s look at a coherent version of Shor error correction, one where there is no measurement, but a set $A$ of 8 ancilla qubits, initially in the $\ket{0}$ state. Their final state, if measured in the computational basis, contains the code syndrome. The error correction is described by a unitary $V$ acting the 9 qubits of the code ($B$) and the 8 ancilla qubits. Since the code corrects both bit-flips and phase-flips, we have:
$$
\begin{align}
V⋅(I\otimes I)\ket{\psi}_B\ket{0}_A&amp;=\ket{\psi}_B\ket{0}_A\\
V⋅(X_k\otimes I)\ket{\psi}_B\ket{0}_A&amp;=\ket{\psi}_B\ket{\phi_{k,X}}_A\\
V⋅(Z_k\otimes I)\ket{\psi}_B\ket{0}_A&amp;=\ket{\psi}_B\ket{\phi_{k,Z}}_A\\
\end{align}
$$
The action of $Y_k=iX_kZ_k$ is also corrected, since it is a phase-flip followed by a bitflip $^*$ :
$$
\begin{align}
V⋅(I_k\otimes I)\ket{\psi}_b\ket{0}_a&amp;=iV⋅(X_k\otimes I)⋅(Z_k\otimes I)\ket{\psi}_b\ket{0}_a\\
&amp;=\ket{\psi}_b\ket{\phi_{k,Y}}_a\\
\end{align}
$$

If we apply this to an arbitrary error $U_k$ on the $k$th qubit, we have :
$$V⋅(U_k\otimes I)\ket{\psi}_b\ket{0}_a=\ket{\psi}_b\left(a\ket0_a+b\ket{\phi_{k,X}}_A+c\ket{\phi_{k,Y}}_A+d\ket{\phi_{k,Z}}_A\right) $$
and the vector $\ket{\psi}_b$ is recovered.