I have a little confusion regarding the following:
$\gamma $ is a piecewise smooth curve from $A$ to $B$ and $h(x,y)$ is a continuously differentiable function on $\gamma$. Let this curve be given by $$t\mapsto (x(t),y(t)),\text{ such that } t\in[a,b] .$$
Then $$\int_{\gamma}dh=\int_{\gamma}\left({{\partial h}\over {\partial x}}dx +{{\partial h}\over {\partial y}}dy\right)=\int_a^b{{\partial h}\over {\partial x}}{{d x}\over {d t}}dt + \int_a^b{{\partial h}\over {\partial y}}{{dy}\over {dt}}dt.$$
Upto this is fine. Next they write: Using chain rule , this becomes $$\int_a^b {{d}\over {dt}}h(x(t),y(t))dt.$$
This step is not clear to me because
$${{d}\over {dt}}[h(x(t),y(t))]={{dh}\over{dt}}|_{(x(t),y(t))}\left({{dx}\over {dt}},{{dy}\over {dt}}\right).$$ How can this be equal to the above?
You got confused by the dimensions. The functions are $$h:\mathbb{R}^2\to\mathbb{R}\\\gamma:\mathbb{R}\to\mathbb{R}^2$$ and if you look closely on the theorem of the chain rule or better yet at the definition of multi-variable differentiation, you'll see how to use it correctly. Just do it step by step:
$$\begin{align}\frac{d}{dt}[h(x(t),y(t))]&=\frac{d}{dt}[h\circ \gamma](t)\\ &=Dh\mid_{\gamma(t)}D\gamma\mid_t\\ &=(\nabla h)^T\mid_{\gamma(t)}\frac{d}{dt}\gamma(t)\\ &=\begin{pmatrix}\frac{\partial h}{\partial x}(\gamma(t)), &\frac{\partial h}{\partial y}(\gamma(t))\end{pmatrix}\begin{pmatrix}\frac{dx}{dt}(t)\\\frac{dy}{dt}(t)\end{pmatrix}\\ &=\frac{\partial h}{\partial x}(\gamma(t))\frac{dx}{dt}(t)+\frac{\partial h}{\partial y}(\gamma(t))\frac{dy}{dt}(t)\end{align}$$
where in the second line I used a simple trick which prevents confusion when using the chain rule. $D$ denotes a general enough differentiation operator, to be specific the Jacobian matrix of which you probably should have heard (?). For a function $f:\mathbb{R}^n\to\mathbb{R}^m$ this is a $m \times n$ matrix. Just start by carefully writing down what the domains and codomains of your functions are, construct the Jacobian and it becomes quite hard to mess up chain rule ever again.