Let $X$ be a normal variety over an algebraically closed field $k$ of characteristic $p>0$ and $S$ be a prime Weil divisor on $X$ which is normal too. Now if $K_X+S$ is NOT $\mathbb{Q}$-Cartier, can I still say that $(K_X+S)|_S=K_S+\text{ Diff}_{S}(0)$ ?

I have looked at the following two books by Janos Kollar: (1) Flips and Abundance for Algebraic Threefolds and (2) Singularities of the Minimal Model Program.

On the 1st book, in Proposition $16.5$ he says the above relation will be true if $\mathcal{O}_X(K_X+S)$ is locally free at all points $P\in S$ such that $\text{codim}_S\ P=1$.

In the 2nd book, Chapter $4$, Page $153$, in a comment before the definition of ''Different II'' he said that the local freeness of $\mathcal{O}_X(K_X+S)$ at codimension $1$ points of $S$ is automatically satisfied if $X$ and $S$ both are normal and I am unable to see this fact!

Since $X$ is normal, $\text{codim }X_{\text{sing}} \geq 2$, so if I choose a codim $1$ point $P\in S$ which is a codim $2$ point of X then I can not really say that $P\in X-X_{\text{sing}}$, so how can I conclude that $\mathcal{O}_X(K_X+S)$ is free at $P$ ? I do realize that this $P$ is in the smooth locus of $S$ since $S$ is normal but that doesn't seem to be helping here!

I am banging my head against this problem, so someone please help me! It's very crucial problem for some other thing that I am trying to prove.

Thanks!