On a scheme $X$, the most general definition of Cartier divisor is a global section in $\Gamma(X, \mathcal{K}^{*}/\mathcal{O}^{*})$, where $\mathcal{K}^{*}$ is the sheaf of invertible elements of the sheaf of total quotient rings.
Alternatively, some texts refer to a Cartier divisor as an equivalence class of pairs $(\mathcal{L}, s)$, where $\mathcal{L}$ is an invertible sheaf on $X$ and $s$ is a non-zero rational section of $\mathcal{L}$.
I have been able to show that these definitions are equivalent in the case that $X$ is a noetherian integral scheme. But how general does this equivalence go? Are they always the same? I expect the problem may occur when you drop reducedness.
Cartier divisors can be defined on any scheme. And to any Cartier divisor $D$ on any scheme $X$ you can associate a line bundle (invertible $\mathcal{O}_X$-module) which is usually denoted by $\mathcal{O}_X(D)$. So there is always a map $$ \begin{align} \mathrm{Div}(X) &\to \mathrm{Pic}(X) \\ D &\to \mathcal{O}_X(D), \end{align} $$ from the group of Cartier divisors to the group of line bundles. This further induces an injective map $\mathrm{DivCl}(X) \to \mathrm{Pic}(X)$, where $\mathrm{DivCl}(X)$ denotes the group $\mathrm{Div}(X)$ modulo linear equivalence. This map is surjective if $X$ is an integral scheme, but for general schemes not all line bundles come from Cartier divisors.
Now to answer your question, for any scheme $X$ and a line bundle $\mathcal{L}$ on it, you have the following natural one to one correspondence $$ \{ \text{effective cartier divisors } D \text{ such that } \mathcal{O}_{X}(D) \cong \mathcal{L} \} \leftrightarrow \{ \text{non-zero divisors of }\Gamma(X, \mathcal{L}) \}\big/\sim, $$ where $ s \sim s'$ if and only if $s' = us$ for some $u \in \Gamma(X, \mathcal{O}_X^{\times}).$ Note that we are only considering effective Cartier divisors, because the associated line bundle of any divisor has a nonzero global section if and only if it is linearly equivalent to an effective one.