I'll suggest here another possible characterisation, expanding on a suggestion of the OP in one of the comments. Again, this is an assertion that certain known properties of line integration characterise it uniquely; this doesn't provide a "new" construction of line integration. Unlike my previous answer, here all the action takes place on the one manifold $M$.
To avoid various technicalities, I'm going to redefine $\mathcal C_M$ to be the set of immersed paths, i.e. smooth paths $c\colon[0,1]\to M$ such that $\dot c(t)\neq0$ for all $t\in[0,1]$. I think this restriction could probably be removed with enough effort.
Theorem:
For any manifold $M$, line integration is the unique function $I\colon\Omega^1(M)\times\mathcal C_M\to\mathbb R$ satisfying the following properties:
- (additivity in the path) Suppose that $c_1$ and $c_2$ are two immersed paths that are composable, i.e. all the derivatives $c_1^{(i)}(1)=c_2^{(i)}(0)$. Then $I(\omega,c_1c_2)=I(\omega,c_1)+I(\omega,c_2)$ for all $\omega\in\Omega^1(M)$. Here $c_1c_2$ denotes the composite path, defined by$$c_1c_2(t)=\begin{cases}c_1(2t)&0\leq t\leq1/2\\c_2(2t-1)&1/2\leq t\leq1.\end{cases}$$
- (additivity in the $1$-form) We have $I(\omega_1+\omega_2,c)=I(\omega_1,c)+I(\omega_2,c)$ for all $\omega_1,\omega_2\in\Omega^1(M)$ and all $c\in\mathcal C_M$.
- (locality) If $\omega\in\Omega^1(M)$ satisfies $\omega_{c(t)}(\dot c(t))=0$ for all $t\in[0,1]$, then $I(\omega,c)=0$.
- (exact forms) If $f\colon M\to\mathbb R$ is smooth, then $I(\mathrm df,c)=f(c(1))-f(c(0))$.
The proof uses two lemmas.
Lemma 1: Let $c$ be an immersed path. Then there is a non-negative integer $N$ such that for all $0\leq k<2^N$, the restriction $c|_{[2^{-N}k,2^{-N}(k+1)]}$ of $c$ to the interval $[2^{-N}k,2^{-N}(k+1)]$ is an embedding.
Proof (outline): This follows from the standard fact that an immersion is locally an embedding (see e.g. this MO question), and that $[0,1]$ is compact.
Lemma 2: Let $c$ be an embedded path in $M$ and $\omega\in\Omega^1(M)$. Then there exists a smooth function $f\colon M\to\mathbb R$ such that $\omega_{c(t)}(\dot c(t))=\mathrm df_{c(t)}(\dot c(t))$ for all $0<t<1$.
Proof: The pullback $c^*\omega$ is a smooth $1$-form on $[0,1]$, hence is $\mathrm df_0$ for some smooth $f_0\colon[0,1]\to\mathbb R$. We want to show that $f_0$ extends to a smooth map $f\colon M\to\mathbb R$ (i.e. $f_0=f\circ c$).
To do this, we first extend $c$ to a smooth map $c\colon(-\epsilon,1+\epsilon)\to M$ for some $\epsilon>0$. This is possible by Borel's Lemma, which says that we can choose smooth maps $(-\epsilon,0]\to M$ and $[1,1+\epsilon)\to M$ having the same higher-order derivatives at $0$ and $1$ as $c$, respectively.
Decreasing $\epsilon$ if necessary, we may even assume that $c\colon(-\epsilon,1+\epsilon)\hookrightarrow M$ is an embedding. The tubular neighbourhood theorem then implies that the embedding $c$ extends to an embedding $\tilde c\colon(-\epsilon,1+\epsilon)\times (-1,1)^{d-1}\hookrightarrow M$, where $d=\dim(M)$. In other words, we have $c(t)=\tilde c(t,0,\dots,0)$ for all $t$.
We now extend $f_0$ as follows. By Borel's Lemma again, we may extend $f_0$ to a smooth function $f_0\colon(-\epsilon,1+\epsilon)\to\mathbb R$, and then extend this again to a smooth function $f_0\colon(-\epsilon,1+\epsilon)\times(-1,1)^{d-1}\to\mathbb R$. Multiplying by an appropriate bump function if necessary, we may assume that $f_0$ vanishes outside $(-\frac12\epsilon,1+\frac12\epsilon)\times(-\frac12,\frac12)^{d-1}$.
We've now constructed an extension $f=f_0\circ\tilde c^{-1}$ of $f$ on the open neighbourhood $\mathrm{im}(\tilde c)$ of the image of $c$. Moreover, we've ensured that this extension has compact support (it vanishes outside a compact subspace), so we can extend $f$ to all of $M$ by specifying that it is $0$ outside $\mathrm{im}(\tilde c)$. This yields the desired $f$. This proves Lemma 2.
Proof of Theorem: We show unicity. Let $I$ and $I'$ be two functions $\Omega^1(M)\times\mathcal C_M\to\mathbb R$ which satisfy the given conditions. We need to show that $I(\omega,c)=I'(\omega,c)$ for all $\omega\in\Omega^1(M)$ and all immersed paths $c$.
To do this, suppose first that $c$ is embedded. By Lemma 2 we can choose a smooth map $f\colon M\to\mathbb R$ such that $\omega_{c(t)}(\dot c(t))=\mathrm df_{c(t)}(\dot c(t))$ for all $c\in[0,1]$. Using additivity in the $1$-form, locality and the condition about exact forms, we find that $I(\omega,c)=I(\mathrm df,c)=f(1)-f(0)$. Since the exact same argument applies to $I'$, we have $I(\omega,c)=I'(\omega,c)$.
Now let us deal with the general case. By Lemma 1 we can choose a non-negative integer $N$ such that $c|_{[2^{-N}k,2^{-N}(k+1)]}$ is an embedded path for all $0\leq k<2^N$. A repeated application of the additivity property implies that $I(\omega,c)=\sum_{k=0}^{2^N-1}I(\omega,c|_{[2^{-N}k,2^{-N}(k+1)]})$ and similarly for $I'$. Since we already know that $I$ and $I'$ agree on embedded paths, we obtain that $I(\omega,c)=I'(\omega,c)$, as desired. This concludes the proof.
Remark:
If one only cares about the integrals of closed 1-forms, then this whole setup can be significantly simplified: one doesn't need to restrict to immersed paths, and one can replace the locality condition above with the more natural condition:
- (locality') If $\omega$ vanishes on an open neighbourhood of the image of $c$, then $I(\omega,c)=0$.
