The first representation of the OP's example repeated:
$$
\mathbf c_1(t) = \left( \frac{1 - (2 - \sqrt2)t - (\sqrt2 - 1)t^2}
{1 - (2 - \sqrt2)t + (2 - \sqrt2)t^2},
\frac{\sqrt2 t - (\sqrt2 - 1)t^2}{1 - (2 - \sqrt2)t + (2 - \sqrt2)t^2} \right)
\quad \mbox{with} \quad 0 \le t \le 1
$$
The second representation of the OP's example repeated:
$$
\mathbf c_2(u) = \left( \frac{1 -u^2}{1 + u^2},
\frac{2u}{1 + u^2} \right) \quad \mbox{with} \quad 0 \le u \le 1
$$
The equality problem for the two representations can be formulated as solving $u$ and $t$ from the equations
$\mathbf c_1(t) = \mathbf c_2(u)$ in such a way that the solutions are a bijection.
Let a computer algebra system (MAPLE)
do the hard work:
c1x(t) := (1-(2-sqrt(2))*t-(sqrt(2)-1)*t^2)/(1-(2-sqrt(2))*t+(2-sqrt(2))*t^2);
c1y(t) := (sqrt(2)*t-(sqrt(2)-1)*t^2)/(1-(2-sqrt(2))*t+(2-sqrt(2))*t^2);
c2x(u) := (1-u^2)/(1+u^2); c2y(u) := 2*u/(1+u^2);
solve(c1x(t)=c2x(u),u);
solve(c1y(t)=c2y(u),u);
solve(c1x(t)=c2x(u),t);
solve(c1x(t)=c2x(u),t);
Throwing away redundant outcomes gives the following end-result.
$$
u = -\frac{(\sqrt2+1)\,t}{t-2-\sqrt2} \quad ; \quad t = \frac{(2+\sqrt2)\,u}{u+1+\sqrt2} \\
\mathbf c_1(t) = \mathbf c_2\left(-\frac{(\sqrt2+1)\,t}{t-2-\sqrt2}\right) \quad ; \quad
\mathbf c_2(u) = \mathbf c_1\left(\frac{(2+\sqrt2)\,u}{u+1+\sqrt2}\right)
$$