Why is the derivative of the real absolute squared different to complex absolute squared

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I know that for $x \in \mathbb{R}$ there is $$ \frac{\mathrm{d}}{\mathrm{d}x} |x|^2 = 2|x| \frac{\mathrm{d}}{\mathrm{d}x} |x| = 2|x| \frac{x}{|x|} = 2x $$ Which makes sense because $|x|^2 = x^2$, but for $z\in\mathbb{C}$ there is $$ \frac{\partial}{\partial z} |z|^2 = \frac{\partial}{\partial z} (z\overline{z}) = \frac{\partial z}{\partial z} \overline{z} + \frac{\partial\overline{z}}{\partial z}z = \overline{z} $$

Both makes sense on their own but shouldn’t they agree on the real part? Why are these two different?

This leads to the second part: which one should I use for $$ \frac{\partial}{\partial f_j} |\langle f_j,f_k \rangle|^2$$ Or does it depend on whether the inner product space is complex or real?

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$$ \begin{align} \frac{\partial }{\partial x} |z|^2 &= \frac{\partial}{\partial x} (x^2 + y^2)\\ &= 2x \end{align} $$

No contradiction. The derivative of $|z|^2$ with respect to $x$ along the real axis agrees with the derivative of $x^2$ with respect to $x$.

The definition of $\frac{\partial f}{\partial z}$ is $\frac{1}{2} (\frac{\partial f}{\partial x} - i \frac{\partial f}{\partial y})$.

This is why the factor of $2$ is disappearing in your computation.

$$ \begin{align} \frac{\partial }{\partial z} |z|^2 &= \frac{1}{2} (\frac{\partial}{\partial x} (x^2 + y^2) - i \frac{\partial }{\partial y} (x^2 + y^2))\\ &= \frac{1}{2} (2x - 2iy)\\ &= x - i y\\ &=\bar{z} \end{align} $$

You might be confused about why we define $\frac{\partial f}{\partial z}$ this way. For that see

https://math.stackexchange.com/a/4529830/34287