$\left(\log _2\left(x\right)-2\right)\left(\log _2\left(x\right)+1\right)<0$
has a solution $\frac{1}{2}<x<4$
But when we take the second part alone that is
$\left(\log _2\left(x\right)+1\right)<0$
it gives a solution $0<x<\frac{1}{2}$ why is $x>\frac{1}{2}$ in the first case but $x<\frac{1}{2}$ in the second case?
To solve this, you need exactly one of $\log_2(x) -2$ and $\log_2(x) + 1$ to be negative. The problem reduces to solving the simultaneous inequalities:
$$\log_2(x) -2 <0 \text{ and } \log_2(x) + 1>0$$
and
$$\log_2(x) -2 >0 \text{ and }\log_2(x) + 1<0$$
to get two solution sets.
So to answer your question, you're correct that $\log_2(x) +1 <0$ when $0<x<\frac12$, but in this region, we also have $\log_2(x) -2 <0$, so overall the inequality is positive.
