First, let's find a rational parametrization for $G$. Dehomogenizing on the affine open set where $X_2 \neq 0$ with coordinates $x = X_0/X_2$ and $y = X_1/X_2$, we find
\begin{align*}
f &= -4x y^2 + 2x^2 + xy + y^2\\
g &= -4xy^2 + 4x^2 + xy - y^2 \, .
\end{align*}
This shows that $(0,0)$ is a double point of $g$ (and $f$). By Bézout's Theorem, a line intersects the curve defined by $G$ in $3$ points, counting multiplicity. Since the origin is a double point of $C_2: G = 0$, then a line through the origin intersects $C_2$ in exactly one other point. This allows us to find a correspondence between the slope of the line and the coordinates of this point of intersection, which yields a parametrization. Below is a picture (made with Sage) of the curve and one line: you can imagine the different points that are traced out as the slope of the line varies.
$\hspace 4.7cm$
Consider the pencil of lines $y = tx$ through the origin with slope $t$. Substituting $y=tx$ into $g$, we have
$$
0 = -4 t^{2} x^{3} - t^{2} x^{2} + t x^{2} + 4 x^{2} = -x^{2} (4 t^{2} x + t^{2} - t - 4) \, .
$$
The factor of $x^2$ corresponds to the double point at the origin, while the second factor is our point of interest. Solving for $x$, we find
$$
4 t^{2} x + t^{2} - t - 4 = 0 \iff x = \frac{-t^2 + t + 4}{4t^2} \, .
$$
Since $y = tx$, this yields the rational parametrization
$$
(x(t),y(t)) = \left(\frac{-t^2 + t + 4}{4t^2}, \frac{-t^2 + t + 4}{4t} \right)
$$
or in projective coordinates
$$
[X_0(T,U) : X_1(T,U) : X_2(T,U)] = [-T^2 U + TU^2 + 4U^3 : -T^3 + T^2 U + 4TU^2 : 4T^2 U] \, .
$$
The above method applies more generally to any degree $d$ curve $C$ with a point $P$ of multiplicity $d-1$. Each line through $P$ intersects $C$ in exactly one other point again by Bézout's Theorem. Taking a pencil of lines through $P$, we obtain a parametrization in the same way as above. For more details, see Franz Winkler's ON SYMBOLIC PARAMETRIZATION OF ALGEBRAIC CURVES or his book with Sendra and Pérez-Diaz, Rational Algebraic Curves: A Computer Algebra Approach, in particular, this page. Another great exercise is to parametrize a conic, like a circle or a hyperbola. Since a conic is smooth and has degree $d = 2$, then every point has multiplicity $1 = d-1$, so we can parametrize it by the method above. This is explored on p. 17 of Cox, Little, and O'Shea's Varieties, Ideals, and Algorithms.
You can use the parametrization above to compute the intersection multiplicities in the way you indicated in the question. However, the only curves that admit rational parametrizations are curves of geometric genus $0$, so you are correct that there are methods for computing intersection multiplicities without finding a rational parametrization. I'll use a method of computing intersection multiplicities that works for plane curves of any genus. In the following, I'll use the notation of $\S3.3$ (p. 36) of Fulton's Algebraic Curves, in particular the axioms he lays out there.
Let's first consider $P_2 = (0:0:1)$, which corresponds to $(x,y) = (0,0)$ in our affine coordinates. Since the lowest degree homogeneous terms in $f$ and $g$ have degree $2$, we immediately see that $P_2$ is a point of multiplicity $2$ on $C_1: F=0$ and $C_2: G=0$. By Axiom (7), we can replace $g$ by
$$
h := f-g = -2x^2 + 2y^2 = 2(y-x)(y+x) \, .
$$
Using the quadratic formula, we find that
$$
2x^2 + xy + y^2 = (y - \alpha x)(y - \overline{\alpha} x)
$$
where $\alpha,\overline{\alpha} = \frac{-1 \pm \sqrt{-7}}{2}$, so $f$ has tangent lines $y - \alpha x$ and $y - \overline{\alpha} x$. Since $f$ and $h$ have no tangents in common, then
$$
I_{P_2}(f,h) = m_{P_2}(f) m_{P_2}(h) = 2 \cdot 2 = 4.
$$
by Axiom (5).
To deal with $P_3 = (1:1:1) = (1,1)$, it is convenient to make the change of coordinates $x \leftarrow x-1, y \leftarrow y-1$ to move $P_3$ to the origin. (Alternatively, you can just write things in terms of $x-1$ and $y-1$.) Under this change of coordinates, we find
\begin{align*}
f &= -4(x+1)(y+1)^2 + 2(x+1)^2 + (x+1)(y+1) + (y+1)^2\\
&= - 4 x y^{2} + 2 x^{2} - 7 x y - 3 y^{2} + x - 5 y\\
h &= 2(y+1)^2 - 2(x+1)^2 = 2 y^{2} -2 x^{2} + 4 y - 4 x \, .
\end{align*}
Since the lowest homogeneous terms appearing in $f$ and $h$ both have degree $1$, we see that $P_3$ has multiplicity $1$ on $C_1$ and $C_2$. Since $f$ and $h$ share no tangents, then
$$
I_{P_3}(f,h) = m_{P_3}(f) m_{P_3}(h) = 1 \cdot 1 = 1
$$
again by Axiom (5).
I'll leave it to you to compute the remaining intersection multiplicities (but feel free to ask if you have difficulty). The computation for $(1:-1:2) = (1/2:-1/2:1)$ should be largely the same: translate to the origin, then check tangents. To deal with the point $(0:1:0)$ we have to change affine charts and instead use the affine open set where $X_1 \neq 0$ with coordinates $u = X_0/X_1$ and $v = X_2/X_1$. In the end, the intersection numbers should sum to $3 \cdot 3 = 9$ again by Bézout's Theorem, which provides a good sanity check.
