How is the magnitude of a direction vector equal to speed? In thomas calculus $12$ edition it gives the position vector along a line as
$\vec r(t) = \vec r_0  + t\vec v$
where $\vec r_0$ is the initial position vector and $\vec v$ is the direction vector.
Then it rewrites it as 
$\vec r(t) = \vec r_0 + t, |\vec v| \, \frac{\vec v}{|\vec v|}$
and it refers to $|\vec v|$ as the speed, which I don't understand. $|\vec v|$ is basically the length of the direction vector. Wouldn't speed be the $y$ component of the direction vector divided by the $x$ component, that is the slope of the position $\vec v$ time graph formed by $\vec r(t)$.

Your staring equation is a vector equation:
$$
\vec r(t)=\vec r_0+\vec v t
$$.
Now: $|\vec v|$ is the modulus of the vector $\vec v$ and this is, by definition what we call the ''speed'' and $\dfrac{\vec v}{|\vec v|}$ is the versor (unitary vector) that represents the direction of  $\vec v$.
So, to write 
$$
\vec r(t)=\vec r_0 +|\vec v|\dfrac{\vec v}{|\vec v|}
$$
is simply a way to separate the speed $|\vec v|$ by the direction $\dfrac{\vec v}{|\vec v|}$.
