You have misunderstood the parallel axis theorem.
The moment of inertia of an object around an axis is equal to
$$I = \iint\limits_R\rho^2\text{d}A$$
where $\rho$ is the distance from any given point to the axis. In the case of a rectangular section around its horizontal axis, this can be transformed into
$$\begin{align}
I_x &= \int\limits_{-b/2}^{b/2}\int\limits_{-h/2}^{h/2}y^2\text{d}y\text{d}x \\
I_x &= \int\limits_{-b/2}^{b/2}\left.\dfrac{1}{3}y^3\right\rvert_{-h/2}^{h/2}\text{d}y\text{d}x \\
I_x &= \int\limits_{-b/2}^{b/2}\dfrac{1}{3}\dfrac{h^3}{4}\text{d}x \\
I_x &= \left.\dfrac{1}{3}\dfrac{h^3}{4}x\right\rvert_{-b/2}^{b/2} \\
I_x &= \dfrac{bh^3}{12}
\end{align}$$
Now, what if we wanted to get the inertia around some other axis at a distance $r$ from our centroid? In this case, all we have to do is:
$$I = \iint\limits_R(\rho+r)^2\text{d}A$$
$$I = \iint\limits_R\left(\rho^2 + 2\rho r + r^2\right)\text{d}A$$
$$I = \iint\limits_R\rho^2\text{d}A + 2r\iint\limits_R\rho\text{d}A + r^2\iint\limits_R\text{d}A$$
The first component $\iint\limits_R\rho^2\text{d}A$ is simply equal to the original moment of inertia. The second component $2r\iint\limits_R\rho\text{d}A$ is equal to zero since we're integrating around the centroid (it'll become a function of $y^2$, which when integrated from $-h/2$ to $h/2$ gives zero). The third component is equal to $Ar^2$. So, in the end, we get:
$$I' = I + Ar^2$$
So, if you want to calculate the moment of inertia of a rectangular section by considering each of its halves (half above the centroid, half below), you need to do:
$$\begin{align}
I_{half} &= \dfrac{b\left(\dfrac{h}{2}\right)^3}{12} \\
I'_{half} &= I_{half} + b\left(\dfrac{h}{2}\right)\left(\dfrac{h}{4}\right)^2 \\
&= \dfrac{bh^3}{96} + \dfrac{bh^3}{32} = \dfrac{bh^3}{24} \\
I_{full} &= 2I'_{half} = \dfrac{bh^3}{12}
\end{align}$$
Which is the original value for the full section. QED.
