So here is my general take, stop me and ask if you have any questions or something flies above your head.
Let's start by going back to the volume of a parallelepiped in ##\mathbb R^3##. As you are certainly familiar with, the volume of the parallelepiped spanned by three vectors ##\vec u##, ##\vec v##, and ##\vec w## is given by the triple product
$$
V(\vec u,\vec v,\vec w) = \vec u \cdot (\vec v \times \vec w)
$$
and is defined as positive whenever the vectors form a right-handed system and negative if they form a left-handed system. In component form in Cartesian coordinates, this translates into
$$
V = \epsilon_{ijk} u^i v^j w^k.
$$
Going to the three-dimensional volume element that we are to integrate in volume integrals in ##\mathbb R^3##, we look for the volume that is within ##\chi_0^a < \chi^a < \chi_0^a + d\chi^a##, where ##\chi^a## are general coordinates (e.g., cylinder or spherical or any other coordinate system). That volume is spanned by the vectors ##(\partial \vec x/\partial \chi^a) d\chi^a## (no sum over ##a##) and therefore the volume element is
$$
dV = V(\vec E_1, \vec E_2, \vec E_3) d\chi^1 d\chi^2 d\chi^3 \equiv \eta_{123} d\chi^1 d\chi^2 d\chi^3,
$$
where ##\vec E_a## are the tangent basis vectors ##\vec E_a = \partial \vec x/\partial \chi^a## and we defined ##\eta_{abc} = V(\vec E_a,\vec E_b,\vec E_c)##. Note that ##\eta_{abc}## is completely asymmetric, but not equal to ##\epsilon_{abc}##. In fact, it is quite straight-forward to show that ##\eta_{abc} = \sqrt{g} \epsilon_{abc}##, where ##g## is the metric determinant.
Now comes the question: What are the properties of the triple product ##V(\vec u, \vec v, \vec w)##? Well, it maps three (tangent) vectors to a number and so it must be a (0,3) tensor. It is also completely anti-symmetric, so it is actually a 3-form (which is just a fancy name for a completely anti-symmetric (0,3) tensor). This is the property that we will use to define integrations on a manifold or on a submanifold of a manifold (which, of course, is a manifold in itself). The integral of a ##k##-form ##\omega## over a ##k##-dimensional submanifold ##S## can be defined according to
$$
\int_S \omega = \int_{S^*} \omega(\dot\gamma_1, \ldots, \dot\gamma_k) dt^1 \ldots dt^k,
$$
where the ##t^i## (##i = 1, \ldots, k##) parametrise ##S##, ##S^*## is the domain of those parameters, and ##\dot\gamma_i## is the tangent vector of the curve of constant ##t^i##. Again, it is rather straight-forward to show that the ##N##-volume element in an ##N##-dimensional manifold with a metric must be of the form ##\sqrt{g} \epsilon_{a_1\ldots a_N}## in order for a set of orthogonal vectors to span the appropriate volume (this is where the ##\sqrt{g}## that you see in all GR action integrals comes from!).
So what happens when you have a ##N-1##-dimensional submanifold that you would like to integrate over? Well, to have the integral properly defined, you need an ##N-1##-form, but what you have access to is the ##N##-volume element so you are essentially missing one of its argument. You can define an ##N-1##-form by already plugging in a vector field into one (typically the first) of its arguments (for an ##N##-form ##\omega## and a vector field ##J##, this would typically be denoted ##i_J\omega##). In a slightly different point of view, if you want to integrate over a ##N-1##-dimensional hypersurface, you plug the ##N-1## tangent vectors into the volume form and you are left with a 1-form so the ##N-1##-hypersurface element is a dual vector - it effectively maps a tangent vector to a number which is the volume spanned by the tangents in the hypersurface and the vector field you are integrating. This is just the flux of the field through the hypersurface as can be glanced in the ##\mathbb R^3## case, where ##d\vec S = [(\partial \vec x/\partial s) \times (\partial\vec x/\partial t)] ds\, dt## and therefore
$$
\vec J \cdot d\vec S = V(\vec J, \partial \vec x/\partial s,\partial \vec x/\partial t) ds\, dt.
$$
With those general preliminaries out of the way, what does this do for you in terms of integrals in spacetime? Well, everything carries over essentially unchanged (although the metric determinant is negative so you replace ##g \to |g|##). The only thing is how you interpret the things physically. The 4-volume element is a 4-form that can be integrated over an entire region of space-time. Depending on the orientation of a 3-hypersurface, you may interpret it differently. If your 3-hypersurface is a simultaneity (however you have defined that), then the surface element is your spatial volume element with a unit normal (this is where the "observer 4-velocity" comes from). This can be seen if you take a simultaneity from a standard coordinate frame in Minkowski space, i.e., ##t = t_0##, then the vectors tangent to your coordinate lines are ##\partial_1##, ##\partial_2##, and ##\partial_3## and therefore
$$
dS_\mu = \epsilon_{\mu 123} dx^1 dx^2 dx^3 = \epsilon_{\mu 123} dV.
$$
Thus, ##J^\mu dS_\mu = J^0 dV = J^\mu U_\mu dV##, where ##U_\mu## is the dual vector related (by the metric) to the 4-velocity ##U^\mu## of an observer at rest in the frame we took the simultaneity surface from. However, this is not the only type of 3-hypersurface you can have. Leaving out other spacelike surfaces such as simultaneities of other frames for the moment, you can have 3-hypersurfaces that have time as a tangent direction, for example the surfaces of constant ##x^3 = x^3_0##. These surfaces would have a 3-hypersurface element given by (order of indices depending on the direction of the surface)
$$
dS_\mu = \epsilon_{\mu 012} dt\, dx^1dx^2 = B_\mu dt\, dA_{12},
$$
where ##dA_{12}## is the area element of the ##x^1##-##x^2##-region integrated over and ##B_\mu## is a dual vector corresponding to picking out the third element of the tangent vector you are contracting with the 3-hypersurface element. In effect
$$
J^\mu dS_\mu = J^3 dt\, dA_{12} = \vec J \cdot d\vec A_{12} dt.
$$
The integral of ##\vec J \cdot d\vec A_{12}## is just the flux through the surface ##x^3 = x^3_0## and integrating it over time gives you the total amount of whatever the current ##J^\mu## is a current of that was carried through the surface over the time interval you integrated over.
This discussion also underlines the role of the 0-component of any current ##J^\mu## as the spatial density of whatever ##J^\mu## is a current of. The zero component is what you integrate over a simultaneity to obtain how much of that thing is carried over that simultaneity into the future.
Now, in GR, your hypersurfaces will typically be more involved, but as long as they are space- or time-like, they can be locally interpreted in a similar manner. If you go back to #2, you can see one application of this type of integrals, the conservation laws. For example, go back to SR again and consider an energy-momentum tensor that is located so that ##T^{\mu\nu} = 0## for sufficiently large spatial coordinates. Then you can integrate the divergence of the energy-momentum tensor over the spacetime volume ##\Omega## between two simultaneities ##\Sigma_1## and ##\Sigma_2## to obtain
$$
\int_\Omega T^{\mu\nu} (\nabla_\mu K_\nu) \sqrt{|g|} d^4 x = \int_{\Sigma_2} T^{\mu\nu} K_\nu dS_\mu - \int_{\Sigma_1} T^{\mu\nu} K_\nu dS_\mu.
$$
As we discussed, the 3-hypersurface elements for these simultaneities are just ##\epsilon_{\mu 123} dV## and so
$$
\int_{\Sigma_2} T^{0\nu} K_\nu dV - \int_{\Sigma_1} T^{0\nu} K_\nu dV = \int_\Omega T^{\mu\nu} (\nabla_\mu K_\nu) \sqrt{|g|} d^4 x.
$$
Letting ##K## be any Killing field means (as discussed in #2) that the right-hand side here vanishes due to the symmetry of ##T^{\mu\nu}##. In particular, letting ##K = \partial_0## would imply that ##K_0 = 1## with all other components vanishing and therefore
$$
\int_{\Sigma_2} T^{00} dV = \int_{\Sigma_1} T^{00} dV,
$$
which is just the conservation of energy. The corresponding argument for the Killing fields ##\partial_i## instead give you the conservation of the momenta.
