What is the hyperbolic angle as a function of $f(t)$ and, in general, two points $(f(t),t)$ and $(f(t+a)$,$t+a)$?
Is the following a valid way to define hyperbolic angle using $f(t)$ and $t$?
Assumptions:
$t>f(t)$,
$\cosh \phi = \frac{t}{\sqrt{t^2-f(t)^2}} = \frac{e^\phi + e^{-\phi}}{2}$,
$\sinh \phi = \frac{f(t)}{\sqrt{t^2-f(t)^2}} = \frac{e^\phi - e^{-\phi}}{2}$,
where $\phi$ is the hyperbolic angle.
Then,
$$\cosh\phi + \sinh\phi =e^\phi = \frac{t+f(t)}{\sqrt{t^2-f(t)^2}}.$$
Taking the $\ln$ of both sides gives
$$\phi(t) = \ln\frac{t+f(t)}{\sqrt{t^2-f(t)^2}}.$$ $$\phi(t) = \ln\sqrt{\frac{t+f(t)}{t-f(t)}}.$$
If the result above isn't valid, then what would the hyperbolic angle as a function of $f(t)$ be?
Note: The assumptions are not negative, I've used "-" as a bullet point.
I'm not sure where this question came from or where this question is going.
But by division, one has $$\tanh\phi=\frac{\sinh\phi}{\cosh\phi}=\frac{f}{t}.$$
So, $$\phi={\rm arctanh}\left(\displaystyle \frac{f}{t}\right).$$
UPDATE:
So, if the curve $(f(t),t)$ describes a [timelike] straight line through the origin event O, then $$f(t)=v t,$$ where $v$ is a constant that could be described as a "slope" (for a spacetime geometry, this would be the "velocity").
Due to the nature of the "rotations" implied by Einstein's postulates (the boosts of the Lorentz group), the "circle" that is invariant under the rotations (boosts) can be described by $(\sinh\phi,\cosh\phi)$, where $\phi$ is an additive parameter which could be associated with the arc-length along the "unit circle" (a unit hyperbola for special relativity).
So, this "slope" $v$ is equal to $\tanh\phi$.
[If we were dealing with Euclidean geometry, there would be instead a circle describable by $(\sin\theta,\cos\theta)$. Then, the slope $v$ of the straight line would equal to $\tan\theta$.]
To derive the "angle measure",
use the definition "circular arc-length divided by the radius of the circle", using the metric of your geometry.
In special relativity (with metric $ds^2=dt^2-dx^2$ and "circle" [the future hyperbola, with spacelike tangents] $R^2=t^2-x^2$), the angle (called the rapidity) is the spacelike-arc length divided by the timelike radial length. $$ \begin{eqnarray*} \phi &=&\frac{1}{R}\int dL\\ &=& \frac{1}{R}\int \sqrt{dx^2-dt^2}\\ &=& \frac{1}{R}\int {dx}{\sqrt{1-\left(\frac{dt}{dx}\right)^2}}\\ &=& \frac{1}{R}\int {dx}{\sqrt{1-\left(\frac{x}{t}\right)^2}}\\ &=& \frac{1}{R}\int {dx}\frac{\sqrt{t^2-x^2}}{t}\\ &=& \frac{1}{R}\int {dx}\frac{R}{\sqrt{R^2+x^2}}\\ &=& \int {dx}\frac{1}{\sqrt{R^2+x^2}}\\ \end{eqnarray*} $$ where we have used that the slope of the tangent line to the "circle" (the hyperbola) $\displaystyle\frac{dt}{dx} = \frac{x}{t}$.
This integrates to ${\rm arcsinh}(x/R)+C$. This is how the hyperbolic function arises from this approach.