Suppose $U\subseteq{\bf R}^2$ is some open region and $f:U\to{\bf R}^3$ defines some smooth surface in space. Now let $\gamma:I\to U$ be a path in $U$ that corresponds to a path $f\circ\gamma:I\to{\cal S}=f(U)\subset {\bf R}^3$ on the surface which we denote by $\cal S$ (here $I$ is an open interval in $\bf R$). How do we find the length of this curve?
The length is computed as
$${\rm length}=\int_I\underbrace{\left\|\frac{d}{dt}(f\circ\gamma)\right\|}_{\large\rm ``speed"}dt=\int_I\left\|J_f(\gamma)\gamma'(t)\right\|dt $$
where $J_F$ is the Jacobian, or matrix of partials, of $f$ with respect to $\gamma$. Note that an expression of the form $\| Jv\|$ may be rewritten as $\sqrt{Jv\cdot Jv}=\sqrt{v^T(J^TJ)v}$. Note that every matrix $A$ determines a bilinear form as $Q_A(u,v)= u^TAv$.
In general, then, suppose we have a manifold $\cal M$ (the device intended to axiomatize curved space), which is essentially a collection of points with some topology; there is no a priori notion of distance between points. To create such a notion of distance, form a collection of bilinear forms $g_p$, one associated to every point $p$ in space, so that any path $\gamma:I\to\cal M$ can be "measured" via
$$\int_I \sqrt{g_\gamma(\gamma',\gamma')}dt.$$
(I am sort of fudging up usual notational conventions for the sake of clarity.) Generally, given charts aka coordinate patches, we represent $g$ as a matrix, and so put two indices under it.
This is the motivation of the metric tensor in differential geometry. Vectors are usually understood to be tangent vectors; there is an abstract object called a "vector space" whose elements are by fiat called "vectors," and to each point in space we attach a vector space called a "tangent space." The derivative $\gamma'$ of a nice path $\gamma:I\to\cal M$ doesn't reside in just one space, it moves along the tangent spaces, which means that $\gamma'(t)$ is always in the tangent space of the point $\gamma(t)\in\cal M$, and furthermore the bilinear form $g_p$ at the point $p\in\cal M$ is always defined on $p$'s tangent space.
There are different, purely algebraic ways of understanding vectors and tensors in the context of abstract algebra (you'll see $\otimes$ symbols everywhere and no Einstein summation notation, for example): these purely algebraic ways are ultimately married to the differential-geometric ways of understanding tensors at the more advanced levels of geometry.
The distance between two points will not change if you change your coordinates. Also the metric tensor itself, as a function, will not change. But the matrix which represents the metric tensor depends on what coordinate system is being imposed on all of the tangent spaces, so the matrix representation of the metric tensor will in fact change with changes of coordinates.
Let me give a simplified overview of the background related to the question. The entire story can be found in the celebrated paper M. Atiyah, R. Bott, V.K. Patodi, "On the heat equation and the index theorem", which is the canonical reference for the subject.
The Riemannian curvature $R_{abcd}$ and the Ricci tensor $R_{ab}$ mentioned in the question are constructed out of the metric in the sense that in a coordinate patch $(U, x^i)$ they are given by a universal formula involving the partial derivatives of the components of the metric.
Explicit expressions can be found, e.g. here.
Moreover, the formulas turn out to be polynomial in the partial derivatives of $g_{a b}$ of all orders $\ge 0$ and the inverse metric $g^{a b}$.
A formula like that gives components of a tensor if they transform under change of coordinates in the tensorial way.
These observations give rise to the following notion.
Let $(M,g)$ be a Riemannian manifold of dimension $n = \dim M$.
Definition 1. A metric invariant (also known as a Riemannian invariant, or an invariant of the Riemannian structure) is a section $P(g)$ of a tensor bundle over $M$, such that for any diffeomorphism $\phi \colon M \to M$ the naturality property holds:
$$
P(\phi^* g) = g^* P(g)
$$
and in any coordinate patch $(U,x^i)$ (that also gives the coordinate trivialization of the tensor bundle where $P(g)$ has values), the components of $P(g)$ are given by a universal polynomial expression in the list of formal variables $\{ g^{i j}, \partial_{k_1} \dots \partial_{k_s} g_{i j} \}$, $s \ge 0$.
Remark 1. If P has values in $\mathbb{R}$, we have a scalar valued invariant.
Remark 2. In a similar fashion we can give a definition of a metric invariant differential operator.
Examples. The metric tensor $g_{a b}$ and its inverse $g^{a b}$, the Riemann curvature tensor $R_{a b}{}^{c}{}_{d}$, the Ricci tensor $R_{a b}$ are tensor valued metric invariants. The scalar curvature $R = g^{a b} R_{a b}$ is a scalar valued tensor invariant. More scalar valued examples are mentioned in another question of the OP. The Levi-Civita connection $\nabla^g$ of the metric $g$ is a metric invariant differential operator. See e.g. Jack Lee, Riemannian Manifolds. An Introduction to Curvature.
As @Jack Lee has pointed in the comments, using the Levi-Civita connection and the Riemannian curvature one can construct many tensor-valued metric invariants, and taking the complete contractions we obtain lots of scalar valued metric invariants. This can be formalized as follows.
Definition 2. A curvature invariant is a linear combination of partial contractions of the iterated covariant derivatives (with respect to the Levi-Civita connection $\nabla$ associated to the metric $g$) of the Riemannian curvature.
More examples can be found here.
The question in the consideration can now be reformulated as follows.
Can all the metric invariants be obtained as curvature invariants?
The answer is known to be positive. This is a consequence of the First Fundamental Theorem of the classical invariant theory. The key geometric tool that is used to reduce this problem to a problem of the representation theory of the orthogonal group is the normal (or geodesic) coordinates. See the details in the aforementioned paper.
This also holds for the metric invariant differential operators.
Best Answer
Since the product rule tells us $0 = \partial( g g^{-1} ) = (\partial g) g^{-1} + g (\partial g^{-1})$, we have a formula for the derivative of the inverse metric:
$$ \partial_l g^{ij} = -g^{ia} g^{jb} \partial_l g_{ab}.$$
Substituting this in to your expression we get
$$ -g^{ia} g^{jb} \partial_l g_{ab} \partial_k g_{ij}.$$
If we swap the dummy indices $a \leftrightarrow i$, $b \leftrightarrow j$ then this is equal to
$$ -g^{ai} g^{jb} \partial_l g_{ij} \partial_k g_{ab};$$
so it's symmetric in $k$ and $l$.