Let $C$ be a nonsingular curve over a field $k$, and let $\delta: C\to C\times C$ be the diagonal (closed) embedding whose image is:
$$\Delta:=\{(x,x)\colon x\in C\}\,.$$
Now $\Delta$ is a divisor in $C\times C$ so we have the line bundle $\mathscr O_{C\times C}(-\Delta)$. I don't understand the following isomorphism of invertible sheaves on $C$:
$$\Omega^1_C\cong \delta^\ast(\mathscr O_{C\times C}(-\Delta))$$
By the theory of differentials we know that $\Omega^1_C\cong \delta^\ast(\mathcal I/\mathcal I^2)$ where $\mathcal I$ is the kernel of the canonical map $$\delta^\#:\mathscr O_{C\times C}\to \delta_\ast \mathscr O_C\,.$$ Now I really don't get why we have the isomorphism $\delta^\ast(\mathscr O_{C\times C}(-\Delta))\cong\delta^\ast(\mathcal I/\mathcal I^2)$.
I think it will be helpful for you to recall the definition of $O(-\Delta)$. You can look in section 14.3 in Ravi's notes. In particular, $O(-\Delta)$ is the ideal sheaf of $\Delta$, which in your set up is $\mathcal I$.
Note moreover, than $O_C$ is (naturally isomorphic to) the quotient of $O_{C \times C}$ by $\mathcal I$... so really what you are using at the end of the day is that affine locally $I \otimes_A A / I = I / I^2 = (I / I^2) \otimes_A A/I$.
(It's been a while since I've thought about this kind of stuff, so be skeptical... you shouldn't trust what anonymous strangers on the internet tell you anyway.)