Integral of a differential 1-form along a curve (clarification on the definition) Let's denote with $(e_1,\dots,e_d)$ the usual basis of $\Bbb R^d$, and with $({e_1}^*,\dots,{e_d}^*)$ the dual basis of its dual space $\Bbb {(R^d)}^*$.
Let $U$ be an open subset of $\Bbb R^d$ and $\omega:U\to \Bbb {(R^d)}^*$ be a $C^\infty$ differential 1-form.
So, by definition, there exist $C^\infty$ functions $\omega_1,\dots,\omega_d:U\to\Bbb R$ such that $\omega=\sum_{j=1}^d\omega_j\,dx^{\,j}$, where each $x^{\,j}$ is the restriction to $U$ of ${e_j}^*$ (so $dx^{\,j}$, the differential of ${x^{\,j}}$, is the constant map $U\to ({\Bbb R^d})^*$, $u\mapsto {e_j}^*$).
Let $\gamma:[a,b]\to U$ be a $C^\infty$ curve.  So, for all $t\in[a,b]$ it is defined $\gamma'(t)$ as an element of $\Bbb R^d$.  Being $\omega(\gamma(t))$ a functional, we can associate to it a vector of $\Bbb R^d$ (its representation with respect to the usual basis, which happens to be the vector $^t(\omega_1(\gamma(t)),\dots,\omega_d(\gamma(t)))\,\,$).
Now, it is licit to take the scalar product $\langle\omega(\gamma(t)),\gamma'(t)\rangle$.
So we define $$\int_\gamma\omega:=\int_a^b\langle\omega(\gamma(t)),\gamma'(t)\rangle\,dt=\int_a^b\biggl(\sum_{j=1}^d\omega_j(\gamma(t))(\gamma^{\,j})'(t)\biggr)dt.$$ and call it the INTEGRAL OF $\omega$ ALONG $\gamma$.
The question is: what is this definition supposed to mean?  What does this integral represent?
 A: You already know what path integration integration means in the special case of a line segment line segment: if your manifold is the real line, $\omega = f(x) \, dx$ and $\gamma$ is any curve given by the identity function on $[a,b]$, then
$$ \int_{\gamma} \omega = \int_a^b f(x) \, dx $$
where the right hand side is the ordinary integral from introductory calculus.
The only thing that matters for an integral over a path is what happens on that path; the definition is essentially just a "change of variable" to convert an arbitrary curve in the original 'variables' into a line segment in the new 'variables'.
IIRC, it is possible to show that the path integral is equal to a suitable Riemann sum, but this approach is more complicated than the change of variable formula, and may stretch the intuition a fair bit. e.g. you have to see the tangent vector $\gamma'(t) \Delta t$ as being a suitable proxy for the displacement from $\gamma(t)$ to $\gamma(t + \Delta t$), which doesn't actually make sense for a general manifold. This is a natural intuition to develop, since in some sense, tangent vectors are supposed to express the intuitive notion of an infinitesimal displacement, but it might not seem natural at first, particularly when you're using them in a setting where displacements are not actually infinitesimal.

You may be confused about another point that is about differential forms and not integrals. Consider the unit circle, drawn in the plane.
$dx$ and $dy$ are differential forms in the plane. Since the circle is a submanifold of the plane, $dx$ and $dy$ are also differential forms on the unit circle.
But on the unit circle, we have $x^2 + y^2 = 1$, and thus we also have $2x\, dx + 2y \, dy = 0$. So if we have a general differential form $f \, dx + g \, dy$ and we 'pull it back' to the circle, we can simplify it to $f \, dx - \frac{gx}{y} \, dy$.
Or, we could note that on the circle, $dx = -\sin \theta \, d \theta$ and $dy = \cos \theta \, d \theta$. So our general differential form would also be equal to
$$ f \, dx + g \, dy = (g \cos \theta - f \sin \theta) \, d \theta $$
Going back to integration, this gives a clear interpretation about what the integral of a differential form along an arc of the circle "ought" to be: e.g. going around the whole circle should give
$$ \int_C f \, dx + g \, dy = \int_C (g \cos \theta - f \sin \theta) \, d \theta = 
\int_0^{2\pi} (g \cos \theta - f \sin \theta) \, d \theta  $$
This corresponds to a different formula:
$$\int_\gamma \omega = \int_{[a,b]} \gamma^*(\omega) = \int_a^b \gamma^*(\omega)\frac{\partial}{\partial x}  \, dx$$
where I've written the evaluation of a differential form at a tangent vector as taking the product: i.e. on the line, any differential form is a multiple of $dx$, and the product is given by
$$\left( f \, dx \right) \frac{\partial}{\partial x} = f $$
This definition is dual to the one you cited. The integrand is formed by combining differential forms on the manifold with tangent vectors on the line. These can be combined in two ways, but both give the same value: your definition pushes the tangent vectors to the manifold before combining them to produce a number. My way pulls the differential form back to the line.
A: A differential 1-form needs a tangent vector to yield a number. Integral over a curve feeds it tangents and sums the results. The pull back gives us a 1-form in the parameter space so we can actually calculate the integral.
