Now that I have a few minutes, let me expand upon the comment I left above.
Surface integral of a vector field
The surface integral of a vector field $v$ along a two-dimensional surface $S\subset \mathbb{R}^3$ has a natural interpretation as a integral of a differential form, and can be easily generalised to arbitrary differentiable manifolds.
First let us recall the formula from multivariable calculus, which expresses the integral as
$$ \int_S v\cdot dS = \int_S (v\cdot n) |dS| $$
(or similarly expressed relative to a parametrization). The key is noting the "dot product" $v\cdot dS$ which is interpreted as the dot product between the vector field $v$ and the unit "outward" normal vector of $S$. What this expression is, really, can be expressed as the following statement about differential forms.
Let $\omega$ be a volume form on some manifold $M$. (So if $M$ has $n$-dimensions, $\omega$ is a differentiable $n$-form.) Via the volume form we can define the notion of volume, and the notion of an integral in the usual way. (I assume you are familiar with that already.) Then the interior derivative $\iota_v\omega$, which is the $n-1$-form defined by
$$ \iota_v\omega(X_2,\ldots,X_n) = \omega(v,X_2,\ldots,X_n) $$
for $v$ a vector field on $M$, is a differentiable form of the top degree when restricted to any $n-1$-dimensional submanifold. And therefore we can integrate it. So in this language, the object $v\cdot dS$ in the surface integral expression is really $\iota_v\omega$, for $\omega$ the usual volume form on $\mathbb{R}^3$, i.e. $dx \wedge dy \wedge dz$. And in this coordinates it is easy to check that what we write as $v\cdot dS$ indeed is equal to $\iota_v\omega$.
Line integral of a vector field
Observe the following: let $\Phi$ be a scalar function on $\mathbb{R}^3$, and let $\gamma:\mathbb{R}\to\mathbb{R}^3$ be a path, then we have the following version of the fundamental theorem of calculus, since $\Phi\circ\gamma$ is a scalar function on $\mathbb{R}$:
$$ \Phi\circ\gamma(b) - \Phi\circ\gamma(a) = \int_a^b \frac{d}{dt}(\Phi\circ\gamma)(s) ds $$
Now, in the right hand side, we can take the chain rule, and get
$$ \Phi\circ\gamma(b) - \Phi\circ\gamma(a) = \int_a^b \nabla \Phi \cdot \gamma'(s) ds $$
Observe that the gradient $\nabla \Phi$ is a vector field. Now this looks somewhat like the definition of integration of a vector field along a curve!
Instead of $\nabla \Phi$, though, we can even get rid of the requirement to have a Riemannian metric (inner product/dot product), by thinking about $d\Phi$, the exterior derivative of $\Phi$, which now is a one form. A one form acts naturally on vectors. So we write
$$ \Phi \circ \gamma(b) - \Phi\circ\gamma(a) = \int_a^b d\Phi(\gamma'(s))ds $$
(On a Riemannian manifold, we have a canonical identification of tangent vectors and cotangent vectors (i.e. vector fields and differentiable one-forms), so it is sometimes used interchangeably. But here we try to abstract away the unnecessary baggage.)
So in a sense, the proper object to consider for line integrals is not a vector field, but a differentiable one-form $\omega$. But now the situation becomes clear(er). You have two possible points of views:
- From the point of view of differential topology, the line integral of a one-form is nothing but the integral of a differentiable form. Remember, you can integrate a differentiable $p$ form over a manifold of dimension $p$. A curve is a manifold of dimension 1, and so over it we can integrate a one-form.
- But we can also think of the alternative characterisation of a line integral via a parametrised curve. And in this sense, you are closer to solving a differential equation: you are looking for a function $F(s)$ defined along the curve $\gamma(s)$, such that $F(a) = 0$ and $F'(s) = \omega(\gamma'(s))$.
Notice that $F(s)$ is only defined along the curve, and generally $F(s)$ may not extend to a function on the manifold $M$ (or $\mathbb{R}^3$). (By this I mean that in general one cannot find a function $G$ defined over $\mathbb{R}^3$ so that for any curve $\gamma$, $F(b) - F(a) = G(\gamma(b)) - G(\gamma(a))$. For this to be true, certain topological assumptions on the underlying domain must be given, and the one-form $\omega$ must be closed ($d\omega = 0$).)
Integrals of a scalar field
Like Zhen mentioned, an abstract way of thinking about them is to think of them as the integral w.r.t. a volume/area/length (differential) form of some scalar function defined along the line or the surface. But there is a difference compared to the previous two cases. In the case of the line integral of a one form, it is the natural integral (a one dimensional integral) of a differential form. If you have any differentiable manifold $M$, a differentiable curve $\gamma$, and some one form $\omega$ defined over $M$ (or just in a neighborhood of $\gamma$), the integral $\int_\gamma \omega$ is defined unambiguously.
In the case of the surface integral of a vector field, you already require one more level of structure. That is, you cannot just count on having a manifold $M$, a codimension-1 submanifold $S$, and the vector field $v$. In addition you must have specified an $n$-form (doesn't necessarily have to be non-vanishing, like a volume form) on $M$ (or at least in a neighborhood of $S$). To repeat what I just said: to "integrate a vector field (aka compute its flux) along a hypersurface", it is necessary that the ambient space itself has some notion of "volume".
In the case of the scalar integral, more information is necessary, if treated as an integral in the sense of differential forms (if you do not just want to choose an arbitrary area form on the surface in an ad hoc manner). Modeling ourselves on the case of the line/surface integrals in Euclidean space, what we are given to start are a manifold $M$, a submanifold $S$ with positive codimension, and a volume form $\omega$ on $M$. To unambiguously define the integral in the sense of differential forms, you need to somehow generate from $M,S,\omega$ an appropriate differential form which you can integrate over $S$.
To make more clear this problem, the point is that for a point $p\in S\subset M$, the tangent space $T_pS$ is well-defined as a subspace of $T_pM$. And while we know that $T_pM / T_pS$ is well-defined as a quotient vector space, there is no canonical way of including $T_pM/ T_pS$ as a vector subspace of $T_pM$ (no preferred choice of basis). Unfortunately, from the natural duality of the tangent and cotangent space, this means that we have the ability to specify which one-forms are orthogonal to $S$, but we don't have the ability to specify which one-forms are parallel to $S$. In short, we are down to the problem in linear algebra that given a just a vector space $U$ and a fixed subspace $V$, there is no canonical way of computing the projection of an arbitrary vector $u\in U$ onto the subspace $V$. And this in turn implies that given a manifold $M$ and a volume form $\omega$ and a submanifold $S$, without further information the notion of a induced volume form on $S$ is not well-defined.
To different degree of specification, there are different ways of overcoming this. (For example, you can prescribe a number of vector fields (same number as the codimension of $S$) along $S$ that are linearly independent and transverse to $S$. By taking the interior derivative of the volume form relative to them you get a form of the correct degree that you can integrate against.) The method used in your multivariable calculus text tends to be one of prescribing a Riemannian metric. A Riemannian metric gives a positive definite inner product on the tangent space of $M$. And thus makes $T_pM$ into an inner product space. Now, on an inner product space the notion of orthogonal projection is well-defined, and so we can have a well defined differential form to integrate over.... well, almost. The problem is that while the Riemannian metric allows you to define an orthonormal basis for the orthogonal complement of $T_pS$ in $T_pM$, it doesn't specify the sign of the basis elements. So instead of an actual differential form, what you have constructed is a differential form up to sign. For the purpose of the actual computation, we normalize the sign of the differential form such that $\int_S dvol$ is positive. This is why line and surface integrals are not-quite integrals relative to differential forms, and thus is parametrization independent (and hence unsigned).
Measure space?
Using the Riemannian metric, you can also interpret scalar integrals on surface or lines as integration w.r.t. measure. Let me just sketch the case of the line for illustration. Let $\gamma:\mathbb{R}\to\mathbb{R}^3$ be a smooth injective curve with some parametrization without critical points. Then the function $|\gamma'|:\mathbb{R}\to\mathbb{R}_+$ is a smooth function (the same is true if you replace $\mathbb{R}^3$ by any Riemannian manifold with smooth metric). Now $\gamma$ also defines a measurable map, so in particular you can pushforward the Lebesgue measure on $\mathbb{R}$ to some measure on $\mathbb{R}^3$, supported on the image of $\gamma$.
You can check that if $\mu$ is a different (smooth) parametrisation of the same curve, then the pushforward measures relative to $\mu$ and $\gamma$ are absolutely continuous relative to each other. And a direct computation shows that the Radon-Nikodym derivative coincides with what you will get by the Jacobian/change-of-variables/chain-rule formula. And so if you somehow choose a parametrisation with unit-speed as privileged, you can use that to define a measure, and then you can think of integration of scalar field as an integration in the sense of measure.
But notice that even in this case, the choice of the privileged measure to integrate against depends on "unit speed" parametrisation, which definition requires having a Riemannian (or Finsler, if you like) metric on the manifold.
