Why doesn't $v_T = \omega r$ involve the direction of its variables? We derived $v_T = \omega r$ by the following procedure, and it's said that $v_T = \omega r$ "is a relation between the magnitudes of the tangential linear velocity and the angular velocity". Why doesn't the formula involve the direction of its variables?

We first define


 A: The direction is implied to be in the $\hat{\phi}$ direction because it is assumed that (eq 8-9) $r$ is not changing with time. Moreover, it's not really implied since it's the tangential velocity and so is in the tangential ($\hat{\phi}$) direction.

To clarify, let's work with an object spinning in a circle in two dimensions, to keep things simple. A point at distance $r$ in the direction $\hat{r}$ is given by
$$\mathbf{r} = r \hat{r}.$$
where $\hat{r} = \cos\phi \hat{x} + \sin\phi \hat{y}$. The time derivative of this point is given by
$$ \mathbf{v} = \frac{dr}{dt} \hat{r} + r \frac{d\phi}{dt}\hat{\phi} $$
By definition $\frac{d\phi}{dt} := \omega$. Now since it is assumed that $dr/dt = 0$ we are left with
\begin{align}
\mathbf{v} &= r \omega \hat{\phi}\\
&=r\omega (-\sin\phi\hat{x} + \cos\phi \hat{y}).
\end{align}
for the full vector description.
A: Suppose we have a rigid object rotating at some angular velocity $\vec{\omega}$ about some fixed axis.  It can be shown that in general, $\vec{v} = \vec{\omega} \times \vec{r}$, where $\vec{r}$ is the instantaneous position of the object.  This more general equation does take the directions of the quantities into account via their cross product.
However, it is common (particularly in introductory physics texts) to consider only the case where $\vec{\omega}$ and $\vec{r}$ are at right angles to each other.  In that case, this equation simplifies to the version you have.  Recall that the magnitude of a cross product is $|\vec{A} \times \vec{B}| = |\vec{A}||\vec{B}|\sin \theta$.  This means that if $\vec{r}$ and $\vec{\omega}$ are perpendicular to each other, then we have $\theta = 90^\circ$ and
$$
v = |\vec{v}| = |\vec{\omega} \times \vec{r}| = |\vec{\omega}||\vec{r}| \sin \theta = \omega r.
$$
