Is there a tangential surface integral? In $\mathbb{R}^2$, we have two different types of line integrals, the tangential line integral
$$\int_C  \mathbf{F}\cdot d\mathbf r$$
and 
the normal line integral
$$\int_C  \mathbf{F}\cdot \mathbf n \,ds.$$
To give a motivation, these two different integrals are very helpful for understanding Divergence theorem and Stoke's theorem.
Let $\mathbf {F} = P \mathbf i + Q\mathbf j$. The Divergence Theorem says
$$\iint_D \text{div}(\mathbf F)\,dx\,dy = \int_C  \mathbf{F}\cdot \mathbf n \,ds.$$
This gives 
$$\iint_D (P_x + Q_y) \,dx\,dy=\int_C \mathbf{F}\cdot (-dy, dx) = \int_C P\,dy - Q\,dx.$$
Stoke's theorem is 
$$\iint_D \text{curl}(\mathbf F)\cdot \mathbf k \,dx\,dy = \int_C  \mathbf{F}\cdot d\mathbf r$$
and this gives 
$$\iint_D (P_y - Q_x) \,dx\,dy=\int_C \mathbf{F}\cdot (dx, dy) = \int_C P\,dx + Q\,dy.$$
Green's theorem can be derived from either of the two above theorems.
However, for all the surface integral, I have only seen the normal surface integral defined by 
$$\iint_D \mathbf F\cdot \mathbf n\, dS$$
is there a similar concept that is close to the tangential line integral $\int_C  \mathbf{F}\cdot d\mathbf r$?
 A: I'd like to say firstly that I have not seen or thought of this concept before, and the definition I provide here does have its flaws. I am posting this not as a definitive answer, but as a foundation for someone who might like to develop this further. 
Given a parametrized surface $T:(s,t)\mapsto(x(s,t),y(s,t),z(s,t))$, there is not a unique tangent vector $\mathbf v$ on the surface, unlike the uniqueness of a unit normal vector up to orientation.
Let's say the surface is defined by $T(s,t) = (x(s,t), y(s,t), z(s,t))$. Then,
$$
DT(s,t) =
\begin{pmatrix}
\frac{\partial x}{\partial s}(s,t) & \frac{\partial x}{\partial t}(s,t) \\
\frac{\partial y}{\partial s}(s,t) & \frac{\partial y}{\partial t}(s,t) \\
\frac{\partial z}{\partial s}(s,t) & \frac{\partial z}{\partial t}(s,t)
\end{pmatrix}
=
\begin{pmatrix}
 \vert&\vert \\
T_s(s,t) & T_t(s,t) \\
 \vert& \vert
\end{pmatrix}.
$$
The vectors $T_s(s,t),T_t(s,t)$ form a basis $B$ for the tangent plane to $T$ at the point $(x(s,t),y(s,t),z(s,t)).$ For a vector $\mathbf v_B$ in the span of $B$, we can express it in terms of the standard basis via $\mathbf v_\text{standard basis} = DT(s,t)\mathbf v_B.$
One way to define this integral, $I_{\mathbf v_B}$, is to pick a unit tangent vector $\mathbf v_B = (c_1,c_2)$ of two components $c_1$ and $c_2$ that we are going to agree to fix in order to specify a set combination of the vectors $T_s$ and $T_t$ (which do not change with respect to each other), and integrate $\mathbf F\cdot DT(s,t)\mathbf v_B$ with respect to $\mathrm d(s,t)$.
Define $I_{\mathbf v_B}$, the integral of a vector field $\mathbf F$ in the direction of $\mathbf v_B$, tangential to the surface $T$, parameters being taken from an open domain $U$ as 
$$
I_{\mathbf v_B} = \iint_U\mathbf F(x(s,t),y(s,t),z(s,t))\cdot D(s,t)\mathbf v_B\,\mathrm d(s,t).
$$
Then, $I_{\mathbf v_B}$ has the geometric interpretation of the net "flow" of $\mathbf F$ over the surface $T$ in a particular direction, tangent to the surface.
The main issue with this definition (assuming I haven't made silly algebra/calculus mistakes) is that instead of specifying a particular direction to integrate through in terms of the standard basis, this definition requires we choose some fixed combination of the surface tangent vectors $T_s$ and $T_t$, which makes it hard to see which direction we are integrating along, I think. I welcome others' thoughts and suggestions for how to improve what I have here.
A: $\newcommand{\Del}{\nabla}\newcommand{\Vec}[1]{\mathbf{#1}}\newcommand{\Reals}{\mathbf{R}}\newcommand{\dd}{\partial}\newcommand{\Brak}[1]{\left\langle #1\right\rangle}$For vector fields $\Vec{F} = P\, \Vec{i} + Q\, \Vec{j}$ in the plane, there is a complex structure, defined by
$$
J(\Vec{F}) = -Q\, \Vec{i} + P\, \Vec{j}.
\tag{1}
$$
If $\Vec{n}$ is an outward-pointing unit normal field along the boundary of a plane region $\dd D$, then $J(\Vec{n}) = \Vec{t}$ is a unit tangent field oriented appropriately for Green's theorem.
If $\Vec{F}$ is a continuously-differentiable vector field in $D$, then
$$
\Del \times J(\Vec{F})
  = \frac{\dd P}{\dd x} + \frac{\dd Q}{\dd y}
  = \Del \cdot \Vec{F}.
\tag{2}
$$
Equation (2) and the fact that
$$
\Brak{\Vec{F}, \Vec{n}}
  = \Brak{J(\Vec{F}), J(\Vec{n})}
  = \Brak{J(\Vec{F}), \Vec{t}}
$$
account for the correspondence between Stokes's theorem and the divergence theorem in the plane.

Fatal snags occur for surfaces in $\Reals^{3}$:


*

*There is no operator on a smooth surface in $\Reals^{3}$ mapping normal vectors to tangent vectors and vice versa. As both G Cab and A Ortiz note, a smooth surface in $\Reals^{3}$ has dimension two but codimension one. Any field $J$ of linear operators cannot be an isomorphism between the normal and tangent spaces.

*There might exist (for a particular surface) a smooth field $J$ of operators sending a unit normal field to a "distinguished" unit tangent field, but no such operator field exists for a general surface: Applying such a $J$ to the outward unit normal field of a sphere, for example, would give a continuous, nowhere-vanishing tangent field on a sphere, which does not exist.

In case it's of interest: The "correct" framework for Stokes's theorem is the exterior calculus of differential forms. Loosely, there is "one theorem for each dimension". In the plane, Stokes's theorem and the divergence theorem are (from this perspective) the same theorem applied to two different differential one-forms: $P\, dx + Q\, dy$, and $-Q\, dx + P\, dy$.
The index-raising and lowering operators associated to the Euclidean metric in $\Reals^{n}$ map vector fields to differential one-forms (associating, e.g., $P\, \Vec{i} + Q\, \Vec{j}$ and $P\, dx + Q\, dy$). This identification allows one to pass between vector and differential form versions of Stokes's theorem in the plane. A one-form, however, is a suitable integrand only on a curve.
To pass between vector and differential form versions of Stokes's theorem on a surface in $\Reals^{3}$, one identifies
$$
\Vec{i} \leftrightarrow dy \wedge dz,\qquad
\Vec{j} \leftrightarrow dz \wedge dx,\qquad
\Vec{k} \leftrightarrow dx \wedge dy.
$$
In fancy language, this is index-lowering followed by the Hodge star-operator.
A: From a dimensional point of view, on a 3D surface, there is a 2D $dS$, and 
if $\mathbf{F} \cdot \mathbf{n}\,dS$ makes sense,  because $\mathbf{F}$ is 
$\text{something}/m^{\,2} $, then $\mathbf{F} \cdot \mathbf{t}\,dS$ makes sense as well.
Of course, on a surface you have two independent tangent vectors. 
A single specific one  will indicate a certain direction and the corresponding 
$\iint\limits_{\text{surface}} {\mathbf{F} \cdot \mathbf{t}\,dS}$ the integral over a narrow band along a path onto the surface.
For example, for a body moving in a viscous fluid we can define $\mathbf{F}$ as the field of the force per unit area exerted by the fluid on the body.
Then $\iint\limits_{\text{surface}} {\mathbf{F} \cdot \mathbf{n}\,dS}$ will give the resultant of the normal pressure,
while $\iint\limits_{\text{surface}} {\mathbf{F} \cdot \mathbf{t}_\mathbf{x} \,dS},\;\iint\limits_{\text{surface}} {\mathbf{F} \cdot \mathbf{t}_\mathbf{y} \,dS}$ will give the resultant of the shear pressure, for example in the $x$ and $y$ direction.
