Do differential equations correspond to vector fields, or differential forms, or both?

Good questions! As far as I see, an ODE system could find its interpretation by either a vector field or a differential form. Let me try to provide a view using fiber bundle. I hope this may partly provide some answer to your questions, and I also very much look forward to further answers from geniuses in our community.

In the following paragraphs, let us focus on the following ODE system $$ \dot{x}^{\mu}(t)=f^{\mu}(t,x^1(t),x^2(t),...,x^n(t)) $$ for $\mu=1,2,...,n$ (I would like to thank @MoisheCohen for suggesting this clarification).

Let $M$ be an $n$-dimensional differentiable manifold, with $\left\{x^{\mu}\right\}_{\mu=1}^n$ being its local coordinates. Let $P=\mathbb{R}\times M$. Consider the following trivial fiber bundle $\pi:P\to\mathbb{R}$, where $\mathbb{R}$ serves as the base space with $t$ being its local coordinate, $M$ plays the role of the fiber, and $P$ becomes the total space.

Consider the tangent bundle of $P$, denoted by $TP$. Obviously, $$ \left\{\frac{\partial}{\partial t},\frac{\partial}{\partial x^{\mu}}:\mu=1,2,...,n\right\} $$ forms a basis of $TP$. Now, we hope to assign a connection on $TP$ by seeking for a direct-sum decomposition $TP=HP\oplus VP$, where $VP$ denotes the vertical space of $TP$, i.e., $\forall\,p=\left(t,x\right)\in\mathbb{R}\times M=P$, $V_pP=T_x(\pi^{-1}(\pi(p))$, while $HP$ is called the horizontal space of $TP$.

Such an assignment could be done in two identical ways. Let us start from an intuitive one. It is obvious that a basis for $VP$ could be $$ \left\{\frac{\partial}{\partial x^{\mu}}:\mu=1,2,...,n\right\}, $$ since each of these vectors is tangential to the fiber. Therefore, it suffices to construct a basis for $HP$. Since $HP$ is a subspace of $TP$, its basis is no more than a linear combination of all base vectors of $TP$. Since $TP=HP\oplus VP$ and the basis for $VP$ does not include $\partial/\partial t$, the basis for $HP$ must include a non-degenerating $\partial/\partial t$ term. Thus a trivial choice of the base vector is $$ \frac{\partial}{\partial t}+f^{\mu}\frac{\partial}{\partial x^{\mu}}, $$ where each $f^{\mu}$ is a function of $t$ and all $x^{\nu}$'s, and the Einstein notation has been employed. To sum up, we have \begin{align} HP&=\text{span}\left\{\frac{\partial}{\partial t}+f^{\mu}\frac{\partial}{\partial x^{\mu}}\right\},\\ VP&=\text{span}\left\{\frac{\partial}{\partial x^{\mu}}:\mu=1,2,...,n\right\}. \end{align} This assignment defines a connection on $P$.

Alternatively, the assignment could also be done by defining a $VP$-valued differential $1$-form on $P$, denoted by $\omega$, such that $HP$ is its kernel and $VP$ is its image. This $\omega$ is easy to be determined. Compatible to the basis for $HP$ and $VP$ from above, it is clear that $$ \omega=\left({\rm d}x^{\mu}-f^{\mu}{\rm d}t\right)\otimes\frac{\partial}{\partial x^{\mu}}. $$ One may check that its kernel is exactly $HP$ defined from above.

Before moving to our ODE system, let us recap the definition of horizontal lift. Let $\gamma:\mathbb{R}\to\mathbb{R}$ be a curve in the base space $\mathbb{R}$. A curve $\sigma:\mathbb{R}\to P$ is called a horizontal lift of $\gamma$ into the total space $P$ if $\pi(\sigma)=\gamma$ and its tangent vector $\dot{\sigma}\in HP$.

Now, we may relate all these arguments to our ODE system. In local coordinates, without loss of generality, take $\gamma(t)=t$. Let its horizontal lift be $\sigma(t)=\left(t,x^1(t),x^2(t),...,x^n(t)\right)$, where each $x^{\mu}=x^{\mu}(t)$ is to be determined. Note that $\dot{\sigma}$ can be taken as a tangent vector by $$ \dot{\sigma}=\frac{\rm d}{{\rm d}t}=\frac{\partial}{\partial t}+\dot{x}^{\mu}\frac{\partial}{\partial x^{\mu}}. $$ Thus the form of base vectors of $HP$, as well as the condition $\dot{\sigma}\in HP$, implies $$ \dot{x}^{\mu}(t)=f^{\mu}(t,x^1(t),x^2(t),...,x^n(t)) $$ for $\mu=1,2,...,n$, which is exactly our ODE system. Alternatively, provided that $\dot{\sigma}\in HP$ is equivalent to $\omega(\dot{\sigma})=0$, we have $$ \left({\rm d}x^{\nu}-f^{\nu}{\rm d}t\right)\biggl(\frac{\partial}{\partial t}+\dot{x}^{\mu}\frac{\partial}{\partial x^{\mu}}\biggr)=0 $$ for $\nu=1,2,...,n$, which also leads to our ODE system $$ \dot{x}^{\nu}(t)=f^{\nu}(t,x^1(t),x^2(t),...,x^n(t)). $$

To sum up, the solution to an ODE system $\dot{x}=f$ can be taken as the horizontal lift of $\mathbb{R}$ (which is the same as the curve in $\mathbb{R}$) into the fiber bundle $P=\mathbb{R}\times M$ whose connection is assigned by $f$, where $x$ is the local coordinate of the fiber $M$. In constructing the connection, $f$ appears to be either the coefficients of the linear combination of tangent vectors, or the coefficients of a differential $1$-form.

That's it! I hope this partly answers your question. And I am looking forward to further answers from geniuses in our community.