Suppose $f: Y \to X$ is a projective birational morphism between two varieties with mild singularities. For example, we can assume $X$ is normal and kawamata log terminal, $Y$ is $\mathbb Q$-factorial. If needed, we can even assume $Y$ is smooth.

My problem is, suppose $L$ is a Cartier divisor on $Y$, which is $\mathbb Q$-linearly equivalent to a pullback of $\mathbb Q$-Cartier divisor (i.e. $mL = f^*D$ for $m \in \mathbb N$, $D$ Cartier on $X$), then when does $L$ a pullback of Cartier divisor?

The case I am aware of is related to the cone theorem. For example, let's consider a normal surface $X$ with minimal resolution $Y \to X$. Suppose $K_Y=f^*K_X$, then in order to show $K_X$ is Cartier, the way I can think of is to evoke the argument for the cone theorem: show $|nK_Y|$ is $f$-globally generated for any $n>m \gg 1$. Then $nK_Y=f^*D_1, (n+1)K_Y=f^*D_2$ with $D_i$ Cartier, thus $K_Y=f^*(D_2-D_1)$ and $K_X=D_2-D_1$ is also Cartier. My feeling is there should be some simple argument in this case!

Are there other criteria (besides the cone theorem) guarantee that $L$ is a pullback of a Cartier divisor? Or what are examples that this property false? (examples in the weaker case where $L$ is $f$-numerically trivial are also very welcome!)

Edit: Here is an example in the book "Birational geometry of algebraic varieties" (Example 1.46) which shows in the $f$-numerically trivial case, the statement false:

Let $E$ be an elliptic curve, set $f: E \times E \to E$, and $\Gamma_0=\{(x,0)\mid x\in E\}, \Gamma_1=\{(x,x) \mid x\in E\}$. Set $L=\Gamma_1-\Gamma_0$. Then $D$ is Cartier. But $D \not\sim_{\mathbb Q} f^*A$ because $D^2=-2$ while $(f^*A)^2=0$.

One can get a similar example where $f$ is birational (see loc. cit).