On Kelvin-Stokes proof without differential forms I was reading a proof of the Kelvin-Stokes theorem (without differential forms) and the first step was defining a Jordan curve $\gamma:[a,b]\rightarrow\mathbb{R}^2$ and a surface $\psi:D\rightarrow\mathbb{R}^3$ where $D$ is the interior of the curve (i.e. the compact portion of $\mathbb{R}^2$). The surface is $S:=\psi(D)$ and the surface boundary is defined $\Gamma(t)=\psi(\gamma(t))$. 
Then, by definition of line integral we have
\begin{align*}
\oint_{\partial S=\Gamma}\vec{F}\cdot\vec{d\vec{\Gamma}} & = \int\limits_{a}^b\big\langle(F\circ\Gamma(t))|\frac{d\Gamma}{dt}(t) \big\rangle dt
\\ & = \int\limits_{a}^b \big\langle(F\circ\Gamma(t))|\frac{d(\psi\circ\gamma)}{dt}(t) \big\rangle dt
\\ & = \int\limits_{a}^b \big\langle(F\circ\Gamma(t))|(J_\psi)_{\gamma(t)}\cdot\frac{d\gamma}{dt}(t) \big\rangle dt
\end{align*}
Where $J_\psi$ is the Jacobian matrix of $\psi$. 
I'm not sure what $(J_\psi)_{\gamma(t)}$ means though, or why it can be used here. Also, the next few lines of the proof are 
\begin{align*}
\big\langle(F\circ\Gamma(t))|(J_\psi)_{\gamma(t)}\cdot\frac{d\gamma}{dt}(t) \big\rangle &  = \big\langle(F\circ\Gamma(t))|(J_\psi)_{\gamma(t)}|\frac{d\gamma}{dt}(t) \big\rangle\
\\ & = \big\langle(^tF\circ\Gamma(t))\cdot(J_\psi)_{\gamma(t)}|\frac{d\gamma}{dt}(t) \big\rangle
\end{align*}
I'm wondering why the "|" symbol is moving around in the interior product, and also what the superscript $t$ in $(^tF)$ means. 
 A: What you have been typing here is certainly not of help in proving, and more important: understanding, Stokes' theorem in ${\mathbb R}^3$. This theorem says that the line  integral of a vector field ${\bf F}$ along the boundary $\partial S$ of a surface is equal to the flux of the derived vector field ${\rm curl}\,{\bf F}$ across $S$.
For the proof of this theorem we have to pull back "everything" to the plane ${\mathbb R}^2$, where we have a simpler version of Stokes' theorem already at our disposal. The surface $S\subset{\mathbb R}^3$ then appears as image of a domain  $\Omega\subset{\mathbb R}^2$, and the boundary $\partial S$ is the image of the Jordan curve $\partial\Omega$. The $2$-dimensional version of Stokes' theorem is Green's theorem
$$\int_{\partial\Omega}(P\>dx+Q\>dy)=\int_\Omega(Q_x-P_y)\>{\rm d}(x,y)\ .\tag{1}$$
You definitely need $(1)$ in the proof of Stokes' theorem, but I see none of it in your question. There are only all sorts of $1$-dimensional integrals in your argument. About these I'd say that they are overloaded with clumsy notation; in particular you use $\cdot$ and $\langle\ |\ \rangle$ at the same time in order to denote the scalar product. We have
$$\int_{\partial S}{\bf F}\cdot d\Gamma=\int_a^b \bigl\langle({\bf F}\circ\Gamma)(t)\bigm|\Gamma'(t)\bigr\rangle\>dt\ .\tag{2}$$
Now by the chain rule
$$\Gamma'(t)=d\psi\bigl(\gamma(t)\bigr).\gamma'(t)\ .$$
Here $d\psi\bigl(\gamma(t)\bigr)\!:\>{\mathbb R}^2\to{\mathbb R}^3$ is a linear map having a $(3\times2)$-matrix $J_{\psi}(\gamma(t))=:J_\psi$. If we write out the integrand on the RHS of $(2)$ in matrix terms we therefore obtain
$$\eqalign{\bigl\langle({\bf F}\circ\Gamma)(t)\bigm|\Gamma'(t)\bigr\rangle&=\bigl[({\bf F}\circ\Gamma)(t)\bigr]^\top\>J_\psi\>[\gamma'(t)]=\bigl(J_\psi^\top\bigl[({\bf F}\circ\Gamma)(t)\bigr]\bigr)^\top \>\bigl[\gamma'(t)\bigr]\cr &=\left\langle J_\psi^\top\bigl[({\bf F}\circ\Gamma)(t)\bigr]\biggm|\gamma'(t)\right\rangle\ .\cr}$$
The ${}^\top$ as well as the superscript ${}^t$ in your source denote matrix transposition. Note that for column vectors $a$, $b$ one has the rule $\langle a|b\rangle=a^\top b$, where on the right hand side we have a matrix product.
