What is the relation between spatial velocity and temporal velocity? Let me define spatial velocity as the velocity at which an object moves through space (just velocity as we all know) and Temporal velocity as the velocity through which it moves through time.
When spatial velocity increases, Temporal velocity decreases.
What is the relation connecting these two quantities?
I think it involves the constant c and some square terms.
I expect a term like
$v^2-t^2=c^2$
where s is spatial velocity,
t temporal velocity
and c ,the speed of light
Update 1 : I found it  $$v^2+(ct)^2 = c^2$$ where v-spatial velocity,
t- temporal velocity,
c-speed of light
Note:Here I have take temporal velocity $t = \frac{dt}{dt'}$.
Let me explain why.
Imagine a space ship traveling at a velocity v.
There is a laser at the bottom of the ship which emits a photon that bounces back from a mirror at the top of the ship. If the time interval needed for the photon to bounce back as observed by a person on earth  the time interval of a process as observed by the earth is dt and the time interval required for the same process to happen, as observed by the space traveller is dt' ,
Then, $$dt' = \frac{dt}{\sqrt{1-\frac{v^2}{c^2}}} \
\hspace{20mm}            
\rightarrow(1)$$
One might be inclined to take the temporal velocity as dt'/dt.
But if we take so, then  dt' is greater than dt implies  the temporal velocity is greater than one.
But if dt' is greater dt that means time is running slowly in the spaceship.
So, by that convection , if time runs slower, Temporal velocity is higher.
Clearly this is not a good convection.
So we can take temporal velocity as $\frac{dt}{dt'}$.
This checks well with our intuitive understanding of temporal velocity and can be considered as a good definition.
So, by this definition and using eqn (1),
$$v^2+(ct)^2=c^2$$
This makes sense, because if an object moves at the speed of light, v=c
so $$c^2+(ct)^2=c^2$$ $$\implies t=0$$
as expected because if an object travels at the speed of light, then its temporal velocity is zero.
So, then I thought It would make more sense to define temporal velocity as ct to make the equation look like
$$v^2+t^2=c^2$$

This means that when an object remains stationary w.r.t us, its spatial velocity is zero and temporal velocity is c. It does not move through space but moves through time at the maximum speed.
Contrary to this, when spatial velocity = c ,then t=0. Which means when an object moves through space at the maximum speed, it can't move through time.
Notice how the sum of the square of spatial velocity and temporal velocity is a constant.
So we are always moving through the spacetime at a constant velocity c.
sometimes the spatial component is higher than temporal component and sometimes the temporal component is greater than spatial component.
One way or the other, the total velocity is same
But then the unit for temporal velocity becomes $m/s^2$
So can I do that?
Can I define temporal velocity as ct?
 A: My UPDATE to this answer is at the end, in response to your updated question.

(My answer tries to use standard terminology, which might help clarify your question. It also fills in more details than you might have wanted. But I do so because there may be related misconceptions.)
The components of the "4-velocity" vector $\tilde V$ (a unit-timelike vector in spacetime)
are
$$\tilde V=(V_t)\hat t +(V_x)\hat x,$$
where
$V_t^2- V_x^2 =1$. This applies only to massive particles.
In terms of rapidity $\theta$, where the "spatial velocity" $v=c\tanh\theta$,
we have
$$\tilde V=(\cosh\theta)\hat t +(\sinh\theta)\hat x.$$

*

*The "temporal component of the 4-velocity" is $V_t=\cosh\theta=\frac{1}{\sqrt{1-(v/c)^2}}=\gamma=\mbox{time dilation factor}$.

*The "spatial component of the 4-velocity" is $V_x=\sinh\theta=\frac{(v/c)}{\sqrt{1-(v/c)^2}}=\displaystyle\frac{v}{c}\gamma=\mbox{"dimensionless proper-velocity"}=\mbox{"dimensionless celerity"}$.
https://en.wikipedia.org/wiki/Proper_velocity (proper-velocity is not a good term. Celerity is better.)

*The ratio of these components is $\displaystyle\frac{V_x}{V_t}=\tanh\theta=\displaystyle\frac{v}{c}
=\mbox{"dimensionless spatial velocity"}$.
This $v$ is the usual "velocity in space" (whose magnitude is always less than $c$ for massive particles)

Note that the usual "spatial velocity [through space]"
is not the same thing as the spatial-component of the 4-velocity [through spacetime].


from my post at
https://www.physicsforums.com/threads/velocity-through-spacetime.957038/post-6068213

So, multiplying the 4-velocity by $c$ to give it dimensions of speed,
$$c\tilde V=(cV_t)\hat t +(cV_x)\hat x,$$
the temporal component is $c\gamma$,
the spatial component is $v\gamma=\mbox{"proper velocity"}=\mbox{"celerity"}$,
the ratio of components is $(v/c)$, 
where
$v=\mbox{"spatial velocity"}$ [the usual velocity in space]

... of the massive particle.
In terms of $\gamma$,
we have
$$c\tilde V=(\gamma c)\hat t +(\gamma v)\hat x.$$
Note that
$$(\gamma c)^2-(\gamma v)^2=\gamma^2(c^2-v^2)=\gamma^2 c^2(1-(v/c)^2)=c^2$$

This $c\tilde V$ could be thought of as the "momentum 4-vector divided by the mass of the massive particle": $\displaystyle\frac{mc\tilde V}{m}=\frac{\tilde P}{m}$
(It does not apply to light.)


UPDATE
Your updated question reminds me of an old calculation of mine
that seems related to your line of questioning.
I will reproduce parts of it here because the original version (at
https://www.physicsforums.com/threads/speed-of-light.75111/#post-565681 )
looks like it has some formatting issues.

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)....

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
$\displaystyle \frac{d\tau}{dt}$
(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 $\theta$:
\begin{align*}
c^2\left(dt/d\tau\right)^2 - \left(d\vec{x}/d\tau\right)^2 &= c^2\\
c^2\left(\displaystyle\frac{(AO)}{(PO)}\right)^2 - \left(c\displaystyle\frac{(PA)}{(PO)}\right)^2 &=\\
c^2(\cosh\theta)^2 - c^2(\sinh\theta)^2 &=\\
\end{align*}
and
\begin{align*}
c^2\left(d\tau/dt\right)^2 + \left(d\vec{x}/dt\right)^2 &= c^2\\
c^2\left(\displaystyle\frac{(PO)}{(AO)}\right)^2 + \left(c\displaystyle\frac{(PA)}{(AO)}\right)^2 &=\\
c^2\left(\displaystyle\frac{1}{\cosh\theta}\right)^2 + c^2(\tanh\theta)^2 &=\\
c^2\left(\displaystyle\frac{1}{\cosh\theta}\right)^2 + 
c^2\left(\displaystyle\frac{\sinh\theta}{\cosh\theta}\right)^2 &=\\
c^2\left(\displaystyle\frac{1+\sinh^2\theta}{\cosh^2\theta}\right) &=
\end{align*}

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
$\color{red}{\mbox{timelike-hypotenuse (PO)}}$ and
the
$\color{blue}{\mbox{spacelike-leg (PA)}}$.
Note carefully that
$\color{red}{\mbox{(PO)}}$ is not Minkowski-perpendicular to $\color{blue}{\mbox{(PA)}}$.

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.

A: @robphy has given a nice and detailed answer, and it's well worth reading and considering. I'm going to throw in a slightly different take on things, namely that you should specify exactly what you mean by "temporal velocity". Generally speaking, spatial velocity of a clock moving with respect to an observer (we consider the observer to be "at rest") along the x-axis is defined as $\frac{dx}{dt}$, where $x$ and $t$ are as measured by the observer. But then what is "temporal velocity"? The obvious analog is $\frac{dt}{dt}$, but this is always 1 and so isn't interesting. Another possibility is $\frac{dt'}{dt}$, where $t'$ is the time as measured by the moving clock. That's a reasonable choice, but note that the original "spatial velocity" used only the observer's coordinates, whereas this uses a mixture of coordinates.
In some sense by comparing $\frac{dx}{dt}$ and $\frac{dt'}{dt}$ we're mixing apples and oranges.
The spatial analog of $\frac{dt'}{dt}$ is $\frac{dx'}{dt}$, but that's not the same as the original definition of spatial velocity.
A: Velocity is all about a ratio: a ratio between a spatial distance and an elapsed time. If you want to consider some other ratio, such as the ratio between time difference on one clock and time difference on another, then I think it would be better not to call this other ratio a 'velocity', but you could call it a 'rate'.
Returning to velocity (i.e. ratio of distance and time), there is relative velocity and another type sometimes called closing velocity which I don't need to get into. The main thing is that if the trajectory in spacetime is a line, then velocity is about the slope of this line. To be precise, it is about the slope of one such line compared to another. But for a given pair of lines there is just one angle between them. So I don't think it is helpful terminology to talk about two velocities (one spatial and one temporal). Rather, there is a single size of the relative velocity between any two worldlines at any given event. The relative velocity is the ratio of the distance over time as one of them moves in the rest frame of the other.
Finally, I think your question is aimed at getting a feel for the way proper time along a worldline increments more and more slowly, compared to time in some inertial frame, as the worldline approaches the speed of light relative to that frame. I would recommend you focus your thinking around the notions of invariant and the related notion of time dilation. By invariant I mean the fact that the combination
$$
c^2 \Delta t^2 - \Delta x^2
$$
has the same value no matter what frame is used to determine the distance $\Delta x$ and the time $\Delta t$ between two given events. That is the thing you want to get your head around, by a plethora of examples of what it tells you about space, time and motion.
