Here's my $0.02 on the "speed through time" and "speed through spacetime" idea (based on my understanding of the transciption of a small passage from the Elegant Universe). (In short, I think my comments below are an elaboration of JesseM's position.)
To me, "speed through spacetime" means the "spacetime-norm of an object's 4-velocity". In relativity, the 4-velocity of an observer is the unit tangent-vector to the observer's worldline. In the attached diagram, it is the Minkowski-unit vectors \hat t and \hat u shown.
Since, by convention, time is measured in seconds and space in meters, one has to multiply the time-components by a speed, (say) "the speed of light" [that is, the spatial-speed of a light ray], in order to work with spacetime components with homogeneous units. Any speed could have been used to make the units homogeneous. In special relativity, it is convenient to use the "speed of light" to avoid the nuisance of carrying around a dimensionless factor.
Now here are some reasons why emphasizing the "speed through spacetime" idea doesn't seem that useful to me.
[As someone must have mentioned before] In special relativity, the "speed through spacetime" of a light ray is ZERO... since light rays have lightlike or "null" tangent-vectors. While certainly true, I wonder if this might be more confusing to a beginner.
Secondly, if one were to formulate a spacetime geometry for Galilean Relativity, one would also normalize 4-velocities as described above. To work with homogeneous units, one would still have to use a speed. Of course, the speed of light wouldn't be anything special in Galilean Relativity. But let's use it (or, if you wish, any other speed playing the role of a conversion unit). One would still say that all observers travel with that same "speed through spacetime". So, the phrase seems rather empty...except to say that "an observer's 4-velocity is a unit vector (in the appropriate spacetime geometry)".
Now, concerning "speed through time"...
If "speed through space" is \displaystyle \frac{dx}{dt} (the rate of change of the object's-spatial-displacement with respect to the object's-time-displacement, \displaystyle\frac{(PA)}{(AO)}=\frac{\sinh\theta}{\cosh\theta}=\tanh\theta, (where \theta is the rapidity), then, by analogy, "speed through time" is \displaystyle \frac{dt}{dt} (the rate of change of the object's-time-displacement with respect to the object's-time-displacement, \displaystyle\frac{(AO)}{(AO)}=1).
Thus, it would seem inappropriate to use "speed through time" to mean \displaystyle <br />
\frac{d\tau}{dt} (the rate of change of the object's-proper-time-displacement with respect to the object's-time-displacement, \displaystyle\frac{(PO)}{(AO)}=\frac{1}{\gamma}=\frac{1}{\cosh\theta}).
"speed through proper-time" might be more appropriate.
Finally, it might be interesting to express the following relations in terms of the rapidity
\theta:
<br />
\begin{align*}<br />
c^2\left(dt/d\tau\right)^2 - \left(d\vec{x}/d\tau\right)^2 &= c^2\\<br />
c^2\left(\displaystyle\frac{(AO)}{(PO)}\right)^2 - \left(c\displaystyle\frac{(PA)}{(PO)}\right)^2 &=\\<br />
c^2(\cosh\theta)^2 - c^2(\sinh\theta)^2 &=\\<br />
\end{align*}<br />
and
<br />
\begin{align*}<br />
c^2\left(d\tau/dt\right)^2 + \left(d\vec{x}/dt\right)^2 &= c^2\\<br />
c^2\left(\displaystyle\frac{(PO)}{(AO)}\right)^2 + \left(c\displaystyle\frac{(PA)}{(AO)}\right)^2 &=\\<br />
c^2\left(\displaystyle\frac{1}{\cosh\theta}\right)^2 + c^2(\tanh\theta)^2 &=\\<br />
c^2\left(\displaystyle\frac{1}{\cosh\theta}\right)^2 + <br />
c^2\left(\displaystyle\frac{\sinh\theta}{\cosh\theta}\right)^2 &=\\<br />
c^2\left(\displaystyle\frac{1+\sinh^2\theta}{\cosh^2\theta}\right) &=<br />
\end{align*}<br />
Note:
In the first expression, it's the usual difference of the square-norms of the Minkowski-perpendicular legs of this triangle. (The signature of the metric is evident here.)
In the second expression (which may appear to look Euclidean at first glance), it is effectively the sum of the square-norms of the timelike-hypotenuse (PO) and the spacelike-leg (PA). Note carefully that (PO) is not Minkowski-perpendicular to (PA).
In summary, the mathematics is, of course, correct.
However, the emphasis on "speed through spacetime" seems empty,
and the use of "speed through time" seems inappropriate since it involves the proper-time of the object along (OP).
Maybe I should get myself a copy of the Elegant Universe to see this passage for myself... in case I've misunderstood the transcription.