I think that I got your mistake! Here it is
$$\left\|\frac{d\bf{v}}{dt} \right\| \ne \frac{d\bf{\left\|v\right\|}}{dt}\tag{1}$$
In fact, this is simply saying that
In general, norm and differentiation are not interchangble.
So we have
$${\bf{v}} \cdot \frac{d\bf{v}}{dt} = \left\|\bf{v}\right\| \left\|\frac{d\bf{v}}{dt}\right\| \cos \theta= \left\|\bf{v}\right\| \frac{d\bf{\left\|v\right\|}}{dt}\tag{2}$$
and hence you can at most cancel $\left\|\bf{v}\right\|$ from the both sides of the equality to get
$$\left\|\frac{d\bf{v}}{dt}\right\| \cos \theta= \frac{d\bf{\left\|v\right\|}}{dt}\tag{3}$$
Now, according to $(1)$, you cannot cancel $\left\|\frac{d\bf{v}}{dt} \right\|$ with $\frac{d\bf{\left\|v\right\|}}{dt}$ and the $\cos\theta$ won't be equal to one.
To get more enlightened, we know that the velocity vector can be written as
$${\bf{v}}=\left\|{\bf{v}}\right\| {\bf{t}}\tag{4}$$
Now, if you take the derivative you will get
$$\frac{d{\bf{v}}}{dt}=\frac{d\left\|{\bf{v}}\right\|}{dt} {\bf{t}}+\left\|{\bf{v}}\right\| \frac{d{\bf{t}}}{dt}\tag{5}$$
Since ${\bf{t}}$ is a unit vector we can conclude
$${\bf{t}} \cdot {\bf{t}} =1 \qquad \to \qquad {\bf{t}} \cdot \frac{d{\bf{t}}}{dt} =0 \tag{7}$$
and consequently
$$\frac{d{\bf{t}}}{dt}
= \left\| \frac{d{\bf{t}}}{dt} \right\| {\bf{n}}
= \left\| \frac{d}{dt} \frac{{\bf{v}}}{\left\|{\bf{v}}\right\|} \right\| {\bf{n}} \tag{8}$$
So the vector $\frac{d{\bf{v}}}{dt}$ can be written as
$$\boxed{
\dfrac{d{\bf{v}}}{dt}=\frac{d\left\|{\bf{v}}\right\|}{dt} {\bf{t}}+\left\|{\bf{v}}\right\| \left\| \frac{d}{dt} \frac{{\bf{v}}}{\left\|{\bf{v}}\right\|} \right\| {\bf{n}}
}
\tag{9}$$
Finally, since ${\bf{t}} \cdot {\bf{n}}=0$ and both are unit vectors, we can compute $\left\|\frac{d\bf{v}}{dt} \right\|$ as follows
$${\left\|\frac{d\bf{v}}{dt} \right\|} = \sqrt{\left[\frac{d\bf{\left\|v\right\|}}{dt}\right]^2 + \left\|{\bf{v}}\right\|^2 \left\| \frac{d}{dt} \frac{{\bf{v}}}{\left\|{\bf{v}}\right\|} \right\|^2} \tag{10}$$
observing Eq.$(10)$, you can see that why $(1)$ is true. There is a special case where the orientation of a vector does not change which means that
$$\mathbf{t}=\frac{{\bf{v}}}{\left\|{\bf{v}}\right\|} = \text{Constant Vector}\tag{11}$$
In that case $(10)$ simplifies to
$$\left\|\frac{d\bf{v}}{dt} \right\| = \left|\frac{d\bf{\left\|v\right\|}}{dt}\right| \tag{12}$$
Best Answer
Velocity is not the derivative of displacement. Rather, it is the derivative of the position vector $r$. The displacement would be something like $dr$. Of course, the velocity is not parallel to the position vector. For example, when a particle moves along the unit circle, the velocity will always be perpendicular to the position vector.
A similar analysis applies to the acceleration. It will be parallel to the change $dv$ in the tangent vector. There is nothing "high level" about this.