$f$ is not ONTO,this is nul homotopic, so it will have fixed point.
let $f:S^1\rightarrow S^1$ be nulhomotopic, then it extends to a map $F$ from $B^2$ to $S^1$ which can be thought of as a map from $B^2$ to $B^2$, and then apply Brouwer's Fixed Point Theorem, $F$ must
have a fixed point,. Since the image of $F$ is contained in $S^1$ so this fixed point must lie in $S^1$ so it is also a fixed point of $f$.
Also we see as this is null homotopic so degree of $f=0$, but there is a result saying $f:S^1\rightarrow S^1$ with no fixed point has degree $(-1)^{n+1}$
You can obtain such a parametrization by starting with something you know and then transforming it to work with your circle.
More concretely, say the circle is the unit circle centered at origin, and say we're going from $A = (1,0)$ to $B = (0,1)$, which sweeps $90^\circ$ clockwise. The parametrization for this would be
$x(S) = \cos(\frac{\pi}{2} S)$
$y(S) = \sin(\frac{\pi}{2} S)$
But in your circle, the parametrization is supposed to sweep counterclockwise, so we can parametrize backwards by replacing $S$ with $1-S$:
$x(S) = \cos(\frac{\pi}{2} (1-S))$
$y(S) = \sin(\frac{\pi}{2} (1-S))$
But this sweeps from $\theta = \frac{\pi}{2}$ to $\theta = 0$, whereas you want $\theta = \frac{3\pi}{4}$ to $\theta = \frac{\pi}{4}$. To adjust for this, we note that this is simply our circle rotated clockwise by $\frac{\pi}{4}$, so add $\frac{\pi}{4}$ to the arguments of sine and cosine in the parametrization:
$x(S) = \cos(\frac{\pi}{2}(1-S) + \frac{\pi}{4})$
$y(S) = \sin(\frac{\pi}{2}(1-S) + \frac{\pi}{4})$
Now all we're missing is the new radius and center of the circle. To adjust for the radius, just scale $x$ and $y$ by the radius:
$x(S) = \frac{1}{\sqrt 2} \cos(\frac{\pi}{2}(1-S) + \frac{\pi}{4})$
$y(S) = \frac{1}{\sqrt 2} \sin(\frac{\pi}{2}(1-S) + \frac{\pi}{4})$
Finally, to translate the circle so that its center is at $(\frac12, -\frac12)$ instead of $(0,0)$, we must add $\frac12$ to $x$ and $-\frac12$ to $y$, so we end up with
$x(S) = \frac{1}{\sqrt 2} \cos(\frac{\pi}{2}(1-S) + \frac{\pi}{4}) + \frac12$
$y(S) = \frac{1}{\sqrt 2} \sin(\frac{\pi}{2}(1-S) + \frac{\pi}{4}) - \frac12$.
You can check that this indeed gives you want you want. The starting point is
$(x(0),y(0)) = (\frac{1}{\sqrt 2} \cos(\frac{3\pi}{4}) + \frac12, \frac{1}{\sqrt 2} \sin(\frac{3\pi}{4}) - \frac12) = (0,0)$
and the endpoint is
$(x(1),y(1)) = (\frac{1}{\sqrt 2} \cos(\frac{\pi}{4}) + \frac12, \frac{1}{\sqrt 2} \sin(\frac{\pi}{4}) - \frac12) = (1,0)$
and moreover our parametrized curve stays on the circle since
$
(x(S) - \frac12)^2 + (y(S) + \frac12)^2 = \frac12 \cos^2(\frac{\pi}{2}(1-S) + \frac{\pi}{4}) + \frac12 \sin^2(\frac{\pi}{2}(1-S) + \frac{\pi}{4}) = \frac12
$
for all $S \in [0,1]$.
Best Answer
You know that functions $\sin$ and $\cos$ are continuous. So the function $g\colon \mathbb R\to \mathbb R^2$ given by $g(x) = (\sin x, \cos x)$ is continuous. Your function is nothing else than a restriction of $g$.
To show that the inverse is not continuous take the sequence $p_n \in \mathbb S^1$ given by $p_n=f(2\pi-1/n)$. Notice that $p_n\to(1,0)$ but $f^{-1}(p_n) \to 2\pi \neq 0 = f^{-1}(1,0)$.