In general, a $(p,q)$-tensor eats $p$ covectors and $q$ vectors and returns a real number; such an object has $p$ indices upstairs and $q$ indices downstairs. The components of the metric tensor are written $g_{ij}$ because the metric $\mathbf g$ is a $(0,2)$-tensor which eats two vectors (and no covectors) and returns their inner product.
If you see the indices written upstairs, then you're not talking about the components of the metric - you're talking about the components of the dual metric $\tilde{\mathbf g}$, which defines an inner product on the space of covectors. The components of the dual metric are the matrix inverse of the components of the metric, which means
$$\tilde g^{ij} g_{jk} = \delta^i_k$$
Conventionally, we drop the tilde and write the components of this object simply as $g^{ij}$, distinguishing it from the metric itself only via placement of the indices.
In Euclidean space and cartesian coordinates, the components of the metric $g_{ij}$ written out in array form are
$$g_{ij} = \pmatrix{1 & 0 &0 &0\\0&1&0&0\\0&0&1&0\\0&0&0&1} \equiv \delta_{ij}$$
My first question is simply this: why do we write $\delta_{ij}$ here, when I am used to seeing the Kronecker delta written as $\delta^i_j$, with one raised index?
Those two symbols have subtly different meanings. They are both equal to $1$ when $i=j$ and $0$ otherwise, but the former transform as the components of a $(0,2)$-tensor (which is to be expected, since it is the metric tensor expressed in a particular choice of coordinates) while the latter transform as the components of a $(1,1)$-tensor (or a linear transformation).
In other aspects of my work, involving quantum field theory, the convention is to write the standard Minkowski metric as $\eta^{ij}$, with two raised indices, as opposed to two lowered indices.
As above, that is the Minkowski dual metric, which is a $(2,0)$-tensor. In cartesian coordinates, the components $\eta^{ij}$ are the same as the components of $\eta_{ij}$, but the same is not true e.g. in spherical coordinates, in which case
$$\eta_{ij} = \pmatrix{-1 &0&0&0\\0&1&0&0\\0&0&r^2&0\\0&0&0&r^2\sin^2(\theta)} \qquad \eta^{ij} = \pmatrix{-1 &0&0&0\\0&1&0&0\\0&0&\frac{1}{r^2}&0\\0&0&0&\frac{1}{r^2\sin^2(\theta)}}$$
Why, in this formula, have we suddenly gone from writing lengths such as $dx^i$ with raised indices to ones such as $x_i$ with lowered ones?
If you go back a step, we have
$$\mathrm ds^2 = |\mathrm d\mathbf x|^2 + K \frac{(\mathbf x\cdot \mathrm d\mathbf x)^2}{a^2-K|\mathbf x|^2}$$
$$ = \delta_{ij} \mathrm dx^i \mathrm dx^j + K \frac{(\delta_{ij} x^i \mathrm dx^j)(\delta_{\ell m} x^\ell \mathrm dx^m)}{a^2 - K|\mathbf x|^2}$$
Relabeling dummy indices on the second term, we have
$$\mathrm dx^2 = \left(\delta_{ij} + K \frac{(\delta_{i\ell} x^\ell \mathrm )(\delta_{jm} x^m \mathrm )}{a^2 - K|\mathbf x|^2}\right) \mathrm dx^i \mathrm dx^j$$
If we use the index-lowering convention $x_i \equiv \delta_{i\ell}x^\ell$, then comparing this expression to the general $\mathrm ds^2 = g_{ij}\mathrm dx^i \mathrm dx^j$ yields the components given in the text. As a side note, it is my personal belief that using the index-lowering convention on the coordinates is not a good idea. It is, however, perfectly well-defined shorthand, so your mileage may vary.
