Let $\pi:\mathbb R\times Q\rightarrow\mathbb R$ be a trivial bundle, local coordinates are $(t,x^i)$, the jet coordinates are written as $(t, x^i,...x^i_{r})$, and for $r=1,2$ let us also use $\dot{x}^i$ and $\ddot{x}^i$. The contact system has local generators $\theta^i_k:=dx^i_k-x^i_{k+1}dt$.
Denote the variational bicomplex as $(\Omega^{p,q},\delta,\overline{d})$, where $p$ is the vertical degree, $q$ is the horizontal degree, $\delta$ and $\overline{d}$ are the vertical and horizontal differentials, respectively. We might also write $\theta^i_k=\delta x^i_k$ as this relation is valid.
Since the base space is one dimensional, $q=0,1$ only. Let $\Phi^k:=\mathscr{I}\Omega^{k,m}$, where $\mathscr{I}$ is the inner Euler operator, so that $\Phi^k\cong H^1(\Omega^{k,\bullet},\overline{d})$. Then $\Phi^1$ is the module of source forms. In local coordinates, a general source form has the expression $$ \Delta = \Delta_i(t,x,\dots,x_r)\delta x^i\wedge dt. $$
Let $\mathfrak{d}:\Phi^k\rightarrow\Phi^{k+1}$ be the induced differential $\mathfrak{d}=\mathscr{I}\circ\delta$, and $\mathfrak{d}$ is also extended via the same formula to $\Omega^{0,1}$.
Recall the standard result that $H^1(\Phi^\bullet,\mathfrak{d})\cong H^2(\mathbb R\times Q)\cong H^2(Q)$ (for simplicity set $\Phi^0:=\mathrm{coker}(\overline{d}:\Omega^{0,0}\rightarrow\Omega^{0,1})$).
The second isomorphism is because $\mathbb R\times Q$ is homotopy equivalent to $Q$, and the projection $\mathbb R\times Q\rightarrow Q$ is such a homotopy equivalence, and we also have that $H^k(\mathbb R\times Q)\cong H^k(J_\infty(\pi))$, where $\pi_{\infty,0}:J_\infty(\pi)\rightarrow\mathbb R\times Q$ is a homotopy equivalence. In short, the ordinary de Rham cohomology of $J_\infty(\pi)$ is isomorphic to that of $Q$.
Then with this understood, let us consider a more or less explicit isomorphism $H^1(\Phi^\bullet,\mathfrak{d})\cong H^2(J_\infty(\pi))$.
Suppose that the source form $\Delta$ is $\mathfrak{d}$-closed (i.e. it satisfies the Helmholtz conditions). This means that $\delta\Delta$ is $\overline{d}$-exact, $\delta\Delta=\overline{d}\Xi$. Now a general $2$-form has the decomposition $\omega = \omega_1+\omega_2$, where $\omega_i\in\Omega^{i,2-i}$, so let us look for an extension $\omega=\Delta+\Theta$, where $\Theta\in\Omega^{2,0}$ such that $\omega$ is $d$-closed.
We have $$ d\omega=\delta\Delta+\overline{d}\Theta+\delta\Theta=\overline{d}(\Xi+\Theta)+\delta\Theta. $$
So let us choose $\Theta:=-\Xi$. Then the first term (of bidegree $(2,1)$) above vanishes.
Note that $\overline{d}\delta\Xi=\delta^2\Delta=0$, so $\delta\Xi\in\Omega^{3,0}$ is $\overline{d}$-closed. But since $\overline{d}$ is exact in positive vertical degree and when the horizontal degree is less than $1$, this implies that $\delta\Xi=0=-\delta\Theta$.
Hence, we get that $\omega:=\Delta-\Xi$ is $d$-closed and thus determines an element of $H^2(J_\infty(\pi))$, which is the sought-after isomorphism in cohomology.
Now let us consider a local section $\gamma:I\rightarrow \mathbb R\times Q$ ($I=[0,1]$ is the closed interval), and a one-parameter deformation $\sigma:I\times I\rightarrow \mathbb R\times Q$ of this local section, i.e. $\sigma(t,0)=\gamma(t)$, and suppose that $\sigma$ is endpoint-preserving, i.e. $\sigma(0,s)=\gamma(0)$ and $\sigma(1,s)=\gamma(1)$. Thus $\sigma$ is a 2-cycle on $\mathbb R$, but since $\gamma$ is a section, it has the form $\gamma(t)=(t,x(t))$ it follows that it is also a 2-cycle in $Q$, moreover any $2$-cycle in $Q$ can be made to have this form.
Now let $X:=\partial\sigma/\partial s$, which is a vertical vector field on $\mathbb R\times Q$ along the section $\sigma(\bullet,s)$.
In coordinates, the pullback $\sigma^\ast\Delta$ takes the form $$ \sigma^\ast\Delta=\Delta_i(t,x(t,s),\dots,x_r(t,s))d(x^i(t,s))\wedge dt = \Delta_i(t,x(t,s),\dots)\frac{\partial x^i}{\partial s}(t,s)ds\wedge dt = (\Delta_i\circ\sigma)X^i ds\wedge dt, $$ where the slight abuses of notations I have committed are hopefully clear (e.g. $x^i(t,s)=(x^i\circ\sigma)(t,s)$).
Now the form $\Theta$ has the general expression $$ \Theta=\frac{1}{2}\sum_{k,l}\Theta_{ij}^{(k,l)}\theta^i_k\wedge\theta^j_l, $$
and we have $$ \sigma^\ast\theta^i_k=\sigma^\ast(dx^i_k-x^i_{k+1}dt)=d(\frac{\partial^k x^i}{\partial t^k})-\frac{\partial^{k+1} x^i}{\partial t^{k+1}}dt=\frac{\partial^{k+1} x^i}{\partial t^k\partial s}ds, $$ so we also have that $\sigma^\ast\Theta=\cdots ds\wedge ds=0$.
Hence, $$ \sigma^\ast\Delta=\sigma^\ast\omega, $$ but we know that $\omega$ is $d$-closed, so the integral has $$ I:= \int_\sigma\Delta=\int_\sigma\omega, $$ and it depends only on the cohomology class $[\omega]$.
Now here we must slightly generalize the question and ask whether $Q$ is allowed to have nontrivial cohomology. If yes, then the first two points in the question are non-equivalent, i.e. the Helmholtz conditions $\mathfrak{d}\Delta=0$ are insufficient in general to write $\Delta$ as an Euler--Lagrange form.
But we can show that the first and the third conditions are equivalent.
Suppose first that $\Delta=\mathfrak{d}L$ for some Lagrangian $L\in\Omega^{0,1}$. Then we have $\delta L=\Delta +\overline{d}\eta$ for some $\eta\in\Omega^{1,0},$ and taking the second vertical differential gives $\delta^2L=\delta\Delta+\overline{d}\delta\eta$, from which we get that $\Xi=-\delta\eta$.
Then if we again set $\omega:=\Delta-\Xi=\Delta+\delta\eta$, we have that $\omega$ is $d$-exact. Specifically, define $\lambda:=L+\eta$, then $$ d\lambda = \delta L -\overline{d}\eta+\delta\eta=\Delta-\Xi=\omega. $$
So then $$ I=\int_\sigma\omega=\int_\sigma d\lambda=\int_{\partial\sigma}\lambda=0, $$ since $\sigma$ is a cycle.
Conversely, assume the third condition, namely that $I=0$ for every cycle $\sigma$. Then it follows from ordinary de Rham theory that $\omega$ is exact. But then it is an easy proof along similar lines that if $\omega=d\lambda$, then the horizontal part of $\lambda$ is a Lagrangian for $\Delta$.
Edit: The answer here is mostly correct up to some subtleties but I am dissatisfied with it. I am leaving it up but start reworking it.
