When is the line integral independent of parameterization? Let $\alpha: [a,b] \rightarrow \mathbb{R}^2$ be a smooth path (i.e. $\alpha'$ is continuous on $[a,b]$), and let $f$ be a continuous vector field. The line integral of $f$ along $\alpha$ is defined as $\int_a^b f[\alpha(t)]\cdot \alpha'(t) dt$. In a typical calculus textbook, it is written as $\int_C f_1 dx + f_2 dy$, suggesting that the line integral depends only on $C$ - the image of the path, not the actual parameterization $\alpha$. 
In general however, this is not true. For example, $\oint_C -ydx + x dy =2\pi$ when integrated along the path $\alpha: [0,2\pi]\rightarrow \mathbb{R}^2, \alpha(t) = (\cos(t), \sin(t))$. The value becomes $4\pi$ when integrated along $\beta: [0,4\pi]\rightarrow \mathbb{R}^2, \beta(t)= (\cos(t),\sin(t))$, even though $\alpha$ and $\beta$ have the same image.
My question: what additional assumption is required to make the notation $\int_C f_1dx +f_2 dy$ unambiguous?
 A: More generally, you can say this:  Let $f:\mathbb{R}^n\rightarrow\mathbb{R}^n$ be a continuous function. Let $\alpha:[0,1]\rightarrow\mathbb{R}^n$ and $\beta:[0,1]\rightarrow\mathbb{R}^n$ be two smooth curves.  Suppose there is a smooth curve $\theta:[0,1]\rightarrow[0,1]$ such that  $\theta(0)=0, \theta(1)=1$, and $\beta(t) = \alpha(\theta(t))$ for all $t \in [0,1]$.   Then: 
$$ \int_0^1 f(\alpha(t))\cdot \alpha'(t)dt = \int_0^1 f(\beta(t))\cdot \beta'(t)dt $$
Proof:  By the chain rule we have $\beta'(t) = \alpha'(\theta(t))\theta'(t)$. Thus: 
\begin{align}
\int_0^1 f(\beta(t))\cdot \beta'(t) dt &= \int_0^1 f(\alpha(\theta(t)))\cdot \alpha'(\theta(t))\theta'(t)dt \\
&= \int_0^1 f(\alpha(u))\cdot \alpha'(u)du
\end{align}
where the second equality is by the change of variables rule for integration of real-valued functions. $\Box$
In your example of a loop traversed once by a curve $\alpha$ and then twice by a curve $\beta$, you cannot write $\beta(t) = \alpha(\theta(t))$ for all $t \in [0,1]$. Defining $\theta(t) = 2t$ does not work because that would only be defined for $t \in [0,1/2]$.   In another example of a clockwise loop versus a counter-clockwise loop, you again cannot find such a $\theta(t)$ function. 

Edit: Some additional observations: 
Allowing for non-injectivity:
The condition on $\theta$ above does not require the functions $\alpha, \beta$, or $\theta$ to be injective.  For example, it works for the following:  Let $\alpha(t)$ be any smooth curve defined over $t \in [0,1]$ (possibly having repeated values over that interval) and define $\beta(t) = \alpha(\theta(t))$, where
$$ \theta(t) = \sin^2\left(\frac{5\pi t}{2}\right) $$
The $\beta$ curve retraces the $\alpha$ curve 5 times, and the first four times cancel each other out.
Path independence when $f$ is a gradient
Suppose there is a smooth function $H:\mathbb{R}^n\rightarrow\mathbb{R}$ such that $f(x) = \nabla H(x)$ for all $x \in \mathbb{R}^n$.  Then path integrals of $f$ depend only on the endpoints of the path, not on the intermediate values of the path.  Indeed: 
$$ \int_0^1 f(\alpha(t))\cdot \alpha'(t) dt = \int_0^1\frac{d}{dt}\left[H(\alpha(t))\right] dt = H(\alpha(1))-H(\alpha(0)) $$
Cauchy integral formula
There are also connections to complex numbers and the Cauchy integral formula. For example, in some cases the path integral of a loopy curve depends only on the number of times the curve winds clockwise around the origin.
A: In general, a line integral of type
$$ I = \int_{C} \sum_{i} F_{x_i} \mathrm{d} x_i, $$
is independent of parameterization when integrand is an exact differential. This means that exists $F$ with total derivative:
$$ \mathrm{d} F = \dfrac{\partial F}{\partial x_1} \mathrm{d} x_1 + \dfrac{\partial F}{\partial x_2} \mathrm{d} x_2 + \dotsb, $$
where, for all variable $x_i$,
$$ F_{x_i} = \dfrac{\partial F}{\partial x_i}. $$
Note that $dF$ must be integrable on $C$.
In fact, let $F \,\colon D\subseteq\mathbb{R^n}\to\mathbb{R}$ be a continuous and differentiable funtion and let $\gamma \,\colon \left[ 0,1 \right] \to D$ be a smooth path or piecewise smooth path (integrable path), then
$$ \int_{\gamma} \mathrm{d}F = \int_{0}^{1} \dfrac{\mathrm{d}F(\gamma(t))}{\mathrm{d}t} \mathrm{d}t = F(\gamma(1)) - F(\gamma(0)). $$

Example of formal proof
Let's assume that $-y dx + x dy$ is an exact differential, then exists $F$ such that
$$ \begin{aligned} 
\mathrm{d} F &= \dfrac{\partial F}{\partial x} \mathrm{d} x + \dfrac{\partial F}{\partial y} \mathrm{d} y
\\
& = -y dx + x dy,
\end{aligned} $$
which implies
$$ \begin{aligned}
\dfrac{\partial F}{\partial x} &= -y,
&
\dfrac{\partial F}{\partial y} &= +x
\end{aligned}. $$
Calculating its partial cross-derivatives,
$$ \begin{aligned}
\dfrac{\partial^2 F}{\partial x \partial y} &= -1,
&
\dfrac{\partial^2 F}{\partial y \partial x} &= +1
\end{aligned}, $$
where
$$ \dfrac{\partial^2 F}{\partial x \partial y} \neq \dfrac{\partial^2 F}{\partial y \partial x} $$
is a contradiction by Schwarz's or Clairaut's theorem. Therefore, $-x dx + y dy$ is not an exact differential, and its line integral is not independent of the chosen parameterization.
