Representing forces as one-forms
To understand what a Newtonian force field is, let's take a look at Newton's second law $$ F = ma $$ This translates to the following differential-geometric relation $$ (m\circ\dot q)^\cdot = F\circ\dot q $$ where $m:\mathrm{T}M\to \mathrm{T}^*M$ maps from velocity to momentum space and $q:I\subset\mathbb R\to M$ is the trajectory.
The force field ends up being a map $$ F:\mathrm{T}M\to \mathrm{T}\mathrm{T}^*M $$
Let $$ \pi:\mathrm{T}^*M\to M \\ \Pi:\mathrm{T}\mathrm{T}^*M\to \mathrm{T}^*M $$ be the bundle projections.
Then $$ m = \Pi\circ F \\ \mathrm{T}\pi\circ F = \mathrm{id}_{\mathrm{T}M} $$
The latter equation is the equivalent of the semi-spray condition and tells us that we're dealing with a second-order field.
Because the bundles $\mathrm{T}\mathrm{T}^*M$ and $\mathrm{T}^*\mathrm{T}M$ are naturally isomorphic - in coordinates, we just switch the components $(x,p;v,f)\mapsto(x,v;f,p)$ - we can represent it as a differential form on $\mathrm{T}M$, which is just the differential $\mathrm dL$ of the Lagrange function (the Euler-Lagrange equation are Newtonian equations of motion).
Now, the space of Newtonian force fields doesn't come with a natural vector space structure, but rather an affine structure. You need to specify a zero force - a force of inertia - to make it into one. Such a force can for example be given by the geodesic spray of general relativity.
Once that's done, you can represent the force field as a section of the pullback bundle $\tau^*(\mathrm{T}^*M)$ where $\tau:\mathrm{T}M\to M$. This is a velocity-dependant covector field, which you can indeed integrate over or derive from a potential function (in case of velocity independence).
Now, for those who are uncomfortable with this level of abstraction, let's try a more hands-on approach:
Geometrically, the acceleration is given by $(x,v;v,a)\in\mathrm{TT}M$. However, that space has the wrong structure - if we add two accelerations acting on the same particle, we end up with $(x,v;2v,a+a')$, which is no longer a valid acceleration.
What we want instead are vectors $(x;a)\in\mathrm{T}M$ or $(x,v;a)\in\tau^*(\mathrm{T}M)$ in case of velocity-dependent accelerations, and a recipe how to get from these to our original acceleration as that's what occurs in our equation of motion.
So let's assume our acceleration is velocity-independent and represented by $(x;a)\in\mathrm{T}M$. By lifting the vector vertically at $(x;v)\in\mathrm{T}M$, we arrive at $(x,v;0,a)\in\mathrm{TT}M$. What's 'missing' is the horizontal component $(x,v;v,0)\in\mathrm{TT}M$.
Even though such a horizontal lift looks trivial in coordinates, it is not a 'natural' operation in differential geometry. You can fix this in two obvious ways by either providing a connection (it's trivial to see how this works out if you take the geometric approach due to Ehresmann) or by manually specifying the 'zero' acceleration due to inertia.
The question that's left to answer is why we're using forces instead of accelerations, or formulated another way, why do we move to the cotangent space?
From the point of view of differential geometry, one answer to that question is because we want to work with potentials, which are less complicated objects, and the differential yields covectors instead of vectors.
Another point of consideration is that $\mathrm{TT^*}M$, $\mathrm{T^*T}M$ and $\mathrm{T^*T^*}M$ are naturally isomorphic, whereas $\mathrm{TT}M$ is not. These isomorphisms lead to several (more or less) equivalent formulations of analytical mechanics, including the Newtonian, Lagrangian and Hamiltonian approach.
Apologies for expanding the scope of the question - feel free to ignore these ramblings ;)
There was a rather lengthy discussion about whether force is naturally a vector or a covector over at physicsforums: http://www.physicsforums.com/showthread.php?t=666861 .
If you define momentum as "that which is conjugate to position," then momentum is a covector. I.e. if you have a Lagrangian, then:
$$p_\mu =\frac{\partial L}{\partial \dot{x}^\mu}$$
Force can then be interpreted as $d p_\mu / d \tau$. Or, you can define force directly from the Lagrangian as:
$$F_\mu=\frac{\partial L}{\partial x^{\mu}}$$
Combined with the argument about work that you provided, where $W=\int F_{i} dx^{i}$, it seems very compelling that force should naturally be interpreted as a covector.
Everything you said is correct. If a force is not conservative then it still makes sense as an 1-form, albeit one that is not exact.
Note also that the condition $\vec{\nabla} \times \vec{F} =0$ for a force to be locally determined by a potential can be written as $d F=0$ so that $F=-d U$ for some function $U$ by the Poincare lemma.
More generally we have p-form potentials $A$ to which we associate p+1-form field strengths $dF$. E.g in electromagnetism (again!) we can combine the vector and scalar potentials into a 1-form on spacetime (3+1=4) and the resulting field strength tensor is this one.