If $(x,y)$ are your standard Cartesian coordinates and $(X,Y,Z)$ are homogeneous coordinates, then $x=X/Z$ and $y=Y/Z$.
So if we have a function $f(x,y)$ we can convert it to a function $F(X,Y,Z)$ by replacing $x$ with $X/Z$ and $y$ with $Y/Z$.
For example, converting the equation of of a circle $x^2+y^2=1$ to homogeneous coordinates would give the equation of a cone $Z^2=X^2+Y^2$.
My question is what is the significance of $F(X,Y,Z)$ in the context of $f(x,y)$?
Note: I don't have a background in projective geometry where homogeneous coordinates are used so I am sorry if my question is trivial in that context.
One thing to remember is that no matter how we write coordinates for the two-dimensional projective plane, it's still two-dimensional.
One way to use three real coordinates to represent a two-dimensional object is to make declare equivalence classes of suitable subsets of $\mathbb R^3 = \{(X,Y,Z) \mid X,Y,Z \in \mathbb R\}$ so that each equivalence class corresponds to a single point of the two-dimensional object. In the case of homogeneous coordinates of the projective plane, the equivalence classes in $\mathbb R^3$ are lines that pass through the origin, $(0,0,0)$, excluding the origin itself. That is, if $(X,Y,Z)$ are homogeneous coordinates for a particular point in the projective plane, then so are $(aX,aY,aZ)$ for any $a \in \mathbb R\setminus\{0\}$. (The coordinates $(0,0,0)$ would be homogeneous coordinates for every projective point if we allowed $a=0,$ so to keep the projective points distinct we have to disallow using $(0,0,0)$ as coordinates of a projective point. Likewise if we did not exclude this point from each line that forms an equivalence class, we would not be able to call those sets equivalence classes.)
So if $F(X,Y,Z)=0$ is the equation in homogeneous coordinates of a figure in the projective plane, we must also have $F(aX,aY,aZ) = 0$ for every $a\in R$. If $f(x,y)=0$ describes a curve in the projective plane, then $F(X,Y,Z)=0$ describes a surface in $\mathbb R^3$ consisting entirely of straight lines through the origin. Your example for $f(x,y)=x^2 + y^2 - 1=0$ illustrates this: the cone $F(X,Y,Z) = X^2 + Y^2 - Z^2 = 0$ is just the union of all lines that pass through $(0,0,0)$ and through some point on the circle $\{(X,Y,Z) \mid X^2 + Y^2 - 1 = 0 \land Z = 1\}.$
In general, if $S$ is a figure in the projective plane satisfying the equation $f(x,y) = 0$, the lines through the origin that satisfy $F(X,Y,Z)=0$ trace a figure in the plane $Z=1$ that is congruent to the finite points of $S$. The lines of $F(X,Y,Z)=0$ that do not intersect the plane $Z=1$ are parallel to $Z=1$ and all lie in the plane $Z=0$; these correspond to the points of $S$ at infinity.
If $f(x,y)=0$ is the equation of the circle $x^2 + y^2 - 1=0$, you get the circle $X^2 + Y^2 - 1 = 0$ in the plane $Z=1$ and you get nothing at all in the plane $Z=0$ (except possibly the point $(0,0,0)$, which is not a set of coordinates for any projective point). If $f(x,y)=0$ is the equation of the parabola $x^2 - y=0$, you get the parabola $X^2 - Y = 0$ in the plane $Z=1$ and the line $X=0$ in the plane $Z=0$.
A difficulty with figures in the projective plane described by just two coordinates is how to represent points that are "at infinity" relative to those two coordinates. That is, we can write $f(x,y)=x-y-1=0$ to describe a straight line, but there is a "hole" in the line at the point at infinity where the two "ends" should connect to each other. The graph described by $F(X,Y,Z)=X-Y-Z=0$ intersects the plane $Z=1$ along the line $X-Y-1=0$, but it also includes the line $X-Y=0$ in the plane $Z=0$. The "line at infinity", which has no points with finite coordinates, is represented by the entire plane at $Z=0$ and nothing else in homogeneous coordinates.