Let $(M^n,g)$ be a $n$ dimensional real Riemannian manifold and $\theta \in TM^{\ast}$ be a 1-form.
Definition. Closed Forms
We say $\theta$ is closed if $d\theta = 0$ a.e.
Definition. Path Independence
Let $p, q$ $\in M^n$ be any two points. Denote all simple curves between $p$ and $q$ by
\[\Omega_{pq} = \{ \alpha: I \longrightarrow M \ \Big| \ \alpha(0) = p \text{ and } \alpha(1) = q \}.\]Then we say a 1-form $\theta \in TM^{\ast}$ is path independent if
\[\alpha \longmapsto \int_{\alpha}\theta = c_{pq}, \text{ some constant}, \ \forall \alpha \in \Omega_{pq}.\]Lemma. $\theta$ Path Independence and Closure Equivalence.
The follow three statements are equivalent.
(A) $\theta$ is closed.
(B) for all simple closed curves $\gamma \subset M^n$,
\[\gamma \longmapsto \int_{\gamma}\theta = 0.\] (C) $\theta$ is path independent.
proof:
($A \Longrightarrow B$ ) Let $\gamma \subset M$ be a simple closed curve. By Stokes theorem,
\[0 = \int_{A} d\theta = \int_{\gamma}\theta, \quad \text{where }A\text{ is the region bounded by }\gamma.\]($B \Longrightarrow C$) Let $\gamma \subset M$ be a simple closed curve. Any closed curve $\gamma$ can be factored into the path product of curves, like $\gamma = \alpha * -\beta$. Here $\alpha$ and $\beta$ are paths between any two points $p, q$ $\in \gamma$. The minus sign in front of $\beta$ simply denotes path $\beta$ parametrized with the opposite orientation, e.g. $-\beta = \beta(-t)$.
Let $p, q \in \gamma$ and $\alpha, \beta \in \Omega_{pq}$ be simple curves. Then $\gamma = \alpha * -\beta$. Since
\[\begin{align} 0 &= \int_{\gamma}\theta = \int_{\alpha * -\beta}\theta = \int_{\alpha}\theta - \int_{\beta}\theta \quad \Longrightarrow \quad \int_{\alpha}\theta = \int_{\beta}\theta = c_{pq}. \end{align}\]$(C \Longrightarrow A)$ $\theta$ is path independent implies
\[\begin{align} \int_{\alpha}\theta = \int_{\beta}\theta = c_{pq} &\quad \Longrightarrow \quad 0 = \int_{\alpha}\theta - \int_{\beta}\theta \\ &\quad \Longrightarrow \quad 0 = \int_{\alpha *-\beta}\theta \\ &\quad \Longrightarrow \quad 0 = \int_{\gamma}\theta \ . \end{align}\]By Stokes Theorem,
\[0= \int_{\gamma}\theta = \int_{A} d\theta, \quad \text{where }A\text{ is the region bounded by }\gamma.\]Since $A$ is arbitrary, $d\theta = 0$ almost everywhere.
In Petersen’s Riemannian Geometry, towards motivating the connection he states but does not prove the following.
Lemma. $X = \nabla f$ if $\theta_X$ is Closed.
If $\theta \in TM^{\ast}$ is closed, then there is a function $f: M \longrightarrow \mathbb{R}$ whose gradient, $\nabla f$, generates $\theta$.
proof:
$\theta = g(X, \cdot)$ for some vector field $X$. By the previous lemma, path independence applies as much to $\theta$ as is it does to $X$. In particular, for any fixed $p \in M$ any $\alpha \in \Omega_{px}$ the function
\[\begin{align} f(x) = c_{px} = \int_{\alpha}\theta & = \int_{\alpha} \langle X, d\alpha \rangle_g, \\ \end{align}\]is well defined. Compute the directional derivative of $f$.
\[\begin{align} \langle \nabla f, v \rangle & = \lim_{t \rightarrow 0} \frac{f(\gamma(t)) - f(x)}{t} \\ & = \lim_{t \rightarrow 0} \frac{1}{t} \Bigg(\int_{\beta_t} \langle X, d\beta_t \rangle_g - \int_{\alpha} \langle X, d\alpha \rangle_g\Bigg) \\ \end{align}\]Here $\gamma(0) = x$ and $\dot\gamma(0) = v$.
Because $X$ is path independent
\[\lim_{t \rightarrow 0} \frac{1}{t} \Bigg(\int_{\beta_t} \langle X, d\beta_t \rangle_g - \int_{\alpha} \langle X, d\alpha \rangle_g\Bigg) = \lim_{t \rightarrow 0} \frac{1}{t}\int_{\gamma} \langle X, d\gamma \rangle_g \\\]Then computing the directional derivative boils down to
\[\begin{align} \langle \nabla f, v \rangle_g & = \lim_{t \rightarrow 0} \frac{1}{t}\int_{\gamma} \langle X, d\gamma \rangle_g \\ \end{align}.\]Define the antiderivative
\[\begin{align} F(s) = \int_{-\epsilon}^{s} \langle X \circ \gamma, d\gamma \rangle_g\\ \end{align}.\]Then the limit in $t$ looks like $F^{\prime}(0)$.
\[\begin{align} \lim_{t \rightarrow 0} \frac{1}{t}\int_{0}^{t} \langle X \circ \gamma, d\gamma \rangle_g &= \lim_{t \rightarrow 0} \frac{F(t) - F(0)}{t} = F^{\prime}(0)\\ \end{align}.\]This is some version of fundamental theorem of calculus.
\[\begin{align} \frac{dF}{dt}\Bigg|_{t=0} &= \frac{d}{dt}\int_{0}^{t} \langle X \circ \gamma, d\gamma \rangle_g \Bigg|_{s=0} \\ &= \langle X \circ \gamma(0), \dot\gamma(0) \rangle_g \\ &= \langle X(x), v \rangle_g \end{align}.\]Therefore at $x \in M$, $\langle \nabla f, v \rangle_g = \langle X(x), v \rangle_g$. For reference, see a proof of the lemma in euclidean space.