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
It is exactly what you wrote in $(3)$. With this, we show that $g_{ij}\,{\rm d}u^i\otimes {\rm d}u^j = g_{ij}\,{\rm d}u^i\,{\rm d}u^j$ as follows: $$g_{ij}\,{\rm d}u^i\otimes\,{\rm d}u^j = \frac{1}{2}(g_{ij}\,{\rm d}u^i\otimes {\rm d}u^j + g_{ji}\,{\rm d}u^j\otimes {\rm d}u^i) = \frac{1}{2} g_{ij}({\rm d}u^i\otimes{\rm d}u^j+{\rm d}u^j\otimes {\rm d}u^i) = g_{ij}\,{\rm d}u^i\,{\rm d}u^j,$$where in the middle $=$ sign we factor out $g_{ji} = g_{ij}$.