So I'm trying to understand the stabilization operation which is to obtain a Heegaard splitting of closed orientable $3$-manifold $M$ of genus $g+1$ from genus $g$.
Given a Heegaard splitting $M = H_g\cup H'_g$ of genus $g$, add an unknotted $1$-handle $B$ to $H_g$ to get a handlebody $H_{g+1}$ of genus $g+1$. Here, we call a handle unknotted if there is a $2$-disk $D$ in $M$ such that $D\cap H_{g+1} = \partial D$ and the curve $\partial D$ goes along $B$ only once, see Figure 1.4
Next, we thicken the disk $D$ to get $C = D\times I$. Note that $B\cup C$ is homeomorphic to a $3$-ball, hence,
$$M\cong H_g\cup (B\cup C)\cup H'_g = (H_g\cup B)\cup (C\cup H'_g),$$
where $H_g \cup B = H_{g+1}$. $\color{\red}{\text{The thickened disk}\ C\ \text{intersects the handlebody}\ H'_g\ \text{in two disks, therefore,}\ C\cup H'_g =H'_{g+1}\ \text{is a handlebody of genus}\ g+1}.$
Why the thickened disk $C$ intersects the handlebody $H_g'$ in two disks?
There is an error in this argument: $ C $ is (by definition) a subset of $ H'_g $, so $ C \cup H'_g = H'_g $ which is not a genus $ g+1 $ handlebody.
There is a step missing in your procedure: deleting the newly added handle $B$ from the ``outer'' handlebody. Your red sentence should read:
The split you end up with is
$$ M = (H_g \cup B) \cup (C \cup H_g' \setminus B) = (H_g \cup B) \cup (H'_g \setminus B) =: H_{g+1}' \cup H'_{g+1} $$
which is disjoint except for the boundaries of the handlebodies.
(Think about taking a handlebody of genus $g$ and drilling out a tunnel that starts and ends at the surface; you have just added another hole, so the result is genus $ g+1 $. The disc $ D $ of your original post is an essential disc in this new handlebody. It isn't used to show the result is of the right genus, but it is used to see that the result is stabilised: take a slice $ D' $ through $ B $, this is an essential disc in the handlebody $ H_{g+1} $, and $ \partial D \cap \partial D' $ intersect transversely in a single point hence the Heegard splitting is stabilised.)