Consider complex function $f(z)=\sqrt{z(z-1)}$, here $z\neq0,1$. I referred to similar questions, found branch cut and branch points. The branch points are $z=0,1$. According to the definition of multivalued function, \begin{equation} \sqrt{z(z-1)}=\vert z(z-1)\vert^{\frac{1}{2}}e^{\frac{i}{2}(\text{arg}(z)+\text{arg}(z-1)+2k\pi)},k=0,1 \end{equation} When the point $z=x+iy, 0<x<\frac{1}{2}, y>0$ is approaching the line $(0,1)$ from above, I want to choose the principal branch i.e. the situation $k=0$. So the expression becomes \begin{equation} \sqrt{z(z-1)}=\vert (x+iy)(x-1+iy)\vert^{\frac{1}{2}}e^{\frac{i}{2}(0+\pi+0)}=\vert (x+iy)(x-1+iy)\vert^{\frac{1}{2}}i \end{equation} And this is $\sqrt{\vert x^2-x\vert}i$ when $y\to0+$.
But when the point $\bar{z}$ approaches the line from below, I found the value is $-\sqrt{\vert x^2-x\vert}i$. The difference $f(x+iy)-f(x-iy)$ is $2\sqrt{\vert x^2-x\vert}i$ when $y\to0+$, however, I tried it in Matlab only to find the opposite answer, which is $-2\sqrt{\vert x^2-x\vert}i$.
What is wrong with my attempt?
Firstly, MATLAB uses as principal argument $-\pi < \operatorname{Arg}(z) \leq \pi$, as do all right-thinking people, so that's what I will do below.
Note the expression you want is $$ \sqrt{z(z-1)} = \sqrt{|z(z-1)|}\exp\left[\frac{\mathrm i}{2} \, \operatorname{Arg}(z(z-1))\right] $$ which is not quite the same as $$ \sqrt{z}\sqrt{z-1} = \sqrt{|z(z-1)|}\exp\left[\frac{\mathrm i}{2} \, \left(\operatorname{Arg} z + \operatorname{Arg} (z-1) \right)\right] $$ If $z = x + \mathrm i y$ then $$ z(z-1) = x(x-1) - y^2 + \mathrm i(2x - 1)y $$ so if $0<x<1/2$ and $y>0$, both the real and imaginary parts are negative. As $y\to 0^+$ then, $\operatorname{Arg}(z(z-1)) \to -\pi$. This sign difference is the explanation for your result.
Incidentally that factor $2x-1$ in the imaginary part means that you get the opposite sign if $1/2<x<1$. This function has a branch cut on the line $x=1/2$.