Quite amusingly, I just realised pretty much this exact question was previously asked on math.SE, and an answer was given there a few years ago... by myself. I completely forgot. I'll copy-paste the answer I gave there and keep the post up for future reference.
Start by considering the following identities, obtained via basic complex algebra and Taylor expansions: $$|\langle \psi,\psi+\delta\psi\rangle|^2 = \|\psi\|^4\left\lvert 1+ \frac{\langle\psi,\delta\psi\rangle}{\|\psi\|^2}\right\rvert^2 = \|\psi\|^4 \left(1 +\frac{2\operatorname{Re}\langle\psi,\delta\psi\rangle}{\|\psi\|^2} + \frac{|\langle\psi,\delta\psi\rangle|^2}{\|\psi\|^4}\right)$$
$$ \|\psi+\delta\psi\|^2=\|\psi\|^2 + \|\delta\psi\|^2 + 2\operatorname{Re}\langle\psi,\delta\psi\rangle $$
$$ \frac{1}{\|\psi+\delta\psi\|^2}\simeq\frac{1}{\|\psi\|^2} \left( 1 - \frac{2\operatorname{Re}\langle\psi,\delta\psi\rangle}{\|\psi\|^2} - \frac{\|\delta\psi\|^2}{\|\psi\|^2} + \frac{4[\operatorname{Re}\langle\psi,\delta\psi\rangle]^2}{\|\psi\|^4} \right) $$
$$ \arccos\sqrt{1-x} \simeq \sqrt x, \quad\text{for } x\to 0^+. $$
Our goal is to compute the quantity $$\gamma(\psi,\psi+\delta\psi) = \arccos\sqrt{\frac{|\langle \psi,\psi+\delta\psi\rangle|^2}{\|\psi\|^2 \|\psi+\delta\psi\|^2} }.$$
Assuming $\|\psi\|^2=1$ for simplicity (we can keep the $\|\psi\|$ terms around but nothing substantial changes in the expressions) we thus have $$ \frac{\lvert\langle\psi,\psi+\delta\psi\rangle\rvert^2}{\|\psi+\delta\psi\|^2} \simeq 1 + \lvert\langle\psi,\delta\psi\rangle\rvert^2 - \|\delta\psi\|^2, $$ therefore $$ \gamma(\psi,\psi+\delta\psi)\simeq\arccos\sqrt{1+\lvert\langle\psi,\delta\psi\rangle\rvert^2-\|\delta\psi\|^2} \simeq \sqrt{\|\delta\psi\|^2-\lvert\langle\psi,\delta\psi\rangle\rvert^2}, $$ and finally $$ ds^2 \equiv \gamma(\psi,\psi+\delta\psi)^2 \simeq \|\delta\psi\|^2-\lvert\langle\psi,\delta\psi\rangle\rvert^2, $$ which is the expression we were looking for.