Yes, the number of orbits is finite. Indeed, as Abdelmalek mentioned, this is the question of classification of pencils of conics. The orbits are the following.
First, assume that two conics in the pencil are non-proportional.

1) Assume that at least one of the conics in the pencil is nondegenerate. Then this conic is isomorphic to $\mathbb{P}^1$ and the intersection points of the pencil is a subscheme of length 4.

1a) If the 4 points are distinct, this is a quadruple of points in general position on $\mathbb{P}^2$, all such quadruples are conjugate, and in appropriate coordinates this pencil can be written as
$$
\langle xy - xz, xz - yz \rangle.
$$

1b) If two points collide, this is a triple a points and a tangent vector at one of them, so the pencil can be written in the form
$$
\langle xy + xz, yz \rangle.
$$

1c) If two pairs of points collide, the pencil can be written in the form
$$
\langle x^2, yz \rangle.
$$

1d) If three points collide, the pencil can be written in the form
$$
\langle x^2 - yz, xy \rangle.
$$

1e) If all four points collide, the pencil can be written in the form
$$
\langle x^2 - yz, y^2 \rangle.
$$

2) Assume now that all conics in the pencil are degenerate. This is possible in either of two cases.

2a) All conics contain a given line. Then each conic is the union of this line and an extra line. Extra lines also form a pencil, and its intersection point can lie on the fixed line or away from it. This gives two more orbits:

2a') $\langle xy, xz \rangle$.

2a'') $\langle xy, x^2 \rangle$.

2b) All conics have a fixed singular point, but no common lines: $\langle x^2, y^2 \rangle$.

3) Next, assume that all conics in the pencil are proportional (equivalently, one of the conics is zero). These orbits are parameterized by the rank of the conic.

3a) $\langle x^2 - yz, 0 \rangle$.

3b) $\langle xy, 0 \rangle$.

3c) $\langle x^2, 0 \rangle$.

3d) $\langle 0, 0 \rangle$.

So, altogether there are 12 orbits. However, I could forget something.

EDIT. The orbit closure order, I think, is the following.

Orbits of type 1 are ordered as $(1e) < (1d), (1c) < (1b) < (1a)$; orbits $(1d)$ and $(1e)$ are incomparable and both sit between $(1e)$ and $(1b)$.

$(2a') < (1d)$, and incomparable with $(1c)$ and $(1e)$.

$(2a'') < (2a')$, $(2a'') < (1e)$.

$(2b) < (1e)$, and incomparable with $(2a')$, $(2a'')$.

$(3a) < (1e)$, and incomparable with type 2 orbits.

$(3b) < (3a)$, $(3b) < (2b)$, $(3b) < (2a'')$.

$(3d) < (3c) < (3b)$.

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