The symbolism $(x,y)$ represents the coordinates of a point. Thus :
$$f(t)=\left(\frac{2c^2\cos(t)}{1+\sin^2(t)},\frac{2c^2\sin(t)\cos(t)}{1+\sin^2(t)}\right)=\left(x(t),y(t)\right)$$
This means that the position of point $(x,y)$ depends on the parameter $t$ : this is the definition of the parametric equation of a curve, where :
$$\begin{cases}
x(t)=\frac{2c^2\cos(t)}{1+\sin^2(t)} \\
y(t)=\frac{2c^2\sin(t)\cos(t)}{1+\sin^2(t)}
\end{cases}$$
Of course, it is necessary to know what is a lemniscate if we want to recognize a lemniscate. One can find this in the mathematical handbooks together with the equation of lemniscate.
For example in http://mathworld.wolfram.com/Lemniscate.html ,
comparing the equations (9), (10) to the above parametric equation, with $a=2c^2$ allows to conclude that the curve is a lemniscate.
At particular values of $t=t_n=\frac{\pi}{2}+n\pi \quad\to\quad \cos(t_n)=0\quad$ hence $f(t_n)=(0,0)$
This means that the curve crosses the point origin $(0,0)$.
For more description of the curve close to the origin, we have to compute the derivatives $x'(t_n)$ and $y'(t_n)$. We find :
$$\begin{cases}
x'(t_n)=-2c^2 \text{ if }n\text{ even , and}\quad =2c^2 \text{ if }n\text{ odd} \\
y'(t_n)=-2c^2
\end{cases}$$
The slope of the curve $\frac{dy}{dx}=\frac{y'}{x'}$ is $1$ for $n$ even and $-1$ for $n$ odd. So, the curve crosses the origin two times respectively with slope $1$ and $-1$.