Does a geometric formulation of Newton's 2nd law require a connection? I've seen stated multiple times that a geometric formulation of Newton's law requires to introduce/define a connection on the space.
For example, a connection is used in this answer to formulate Newton's equation as an "equation of vector fields". The same is hinted at in The foundations of geometric formulation of Newton's axioms, as well as in the Wikipedia page discussing the Newton-Cartan theory.
I don't fully understand this. Naively, I could think that Newton's equation could be stated in general geometric terms using just the differential. In other words, saying something like: $dq(t)\equiv(q(t),\dot q(t))$ is the solution to Newton's equation for a force field $F\equiv F(q,v)$ iff $t\mapsto dq(t)$ is the integral curve of the vector field $\tilde F:TQ\to TTQ$ defined locally by $\tilde F(x,v)\equiv (x,v;v,F(x,v))\in T_{(x,v)}TQ$, where $Q$ is the relevant manifold of configurations.
This definition doesn't seem to require making any reference to a connection, so why is a connection often defined when discussing Newton's equation in geometric terms?
A partial answer might lie in this answer, where they write:

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.

A similar discussion of how a connection provides a "splitting" on the double tangent bundle (or more generally, on the tangent bundle of a vector bundle) is given in this MathOverflow answer.
Still, I'm not clear as to why we should care about this sort of splitting when discussing Newton's second law in this context.
 A: The relevant point is the step where you write $\bar{F}(x,v) = (x,\color{blue}{v};\color{red}{v},F(x,v))$ - you have to think about how the third coordinate (highlighted in red) comes about: While the blue $v$ is $\color{blue}{v}\in T_x Q$, the red is just a component of $(\color{red}{v},a)\in T_{x,v}TQ$ - how do you say what "part" of the vector that component is, and what does it mean for the red and blue $v$ to "both have the value $v$"?
The intention is clearly that there is a split of $T_{x,\color{blue}{v}}TQ = V_{x,\color{blue}{v}} \oplus H_{x,\color{blue}{v}}$ into two parts - the "vertical" $V_{x,\color{blue}{v}}$ where $a$ lives and the horizontal $H_{x,\color{blue}{v}}$ where the $\color{red}{v}$ lives, and an isomorphism $T_xQ\cong H_{x,\color{blue}{v}}$ in order for "both have the value $v$" to be meaningful. Getting the vertical part is easy: It's just the kernel of the natural projection $\mathrm{d}\pi : TTQ\to TQ$ obtained as the derivative of the projection $\pi : TQ\to Q$. But there's no good natural candidate for $H_{x,\color{blue}{v}}$ so this is additional data you need to supply. So in order to write down your $\bar{F}$ you need to define this split at every point, and that is exactly what an Ehresmann connection is.
If your base space $Q$ is a Riemannian manifold, then the associated Levi-Civita connection is a natural choice for the required connection.
A: I would have replied in ACuriousMind's comment section, but it's a little too long for it.
In your post, and subsequently in @ACuriousMind's comments, all you've done is started with a vector field $F:TQ\to TTQ$, and gave a local definition of $\widetilde{F}$. Your $\tilde{F}$ is not defined on the whole of $TQ$, and that's the entire problem. Also, in your comment you ask "is choosing a chart for $TTQ$ somehow already implicitly picking a connection/splitting somehow?", the answer is no. A choice of chart is not the same as a choice of connection. By choosing a chart, sure you can define a connection defined on that domain, i.e on $TU$, simply by "transport of structure" of the flat connection on $T\phi[U]\subset \Bbb{R}^{2n}$. But this is definitely not the same thing as having a connection on entire space $TQ$.
Let's be more explicit about where/how your definition goes wrong. Start with a vector field $F:TQ\to TTQ$ on $TQ$. Choose a chart $(U,\phi)$ and lift this up to corresponding tangent bundles. Then, we get the local representation $F_{\phi}:=T^2\phi\circ F\circ T\phi^{-1}:\phi[U]\times\Bbb{R}^n\to (\phi[U]\times\Bbb{R}^n)\times (\Bbb{R}^n\times\Bbb{R}^n)$. We can write this in the form of
\begin{align}
F_{\phi}(x,u)&= \bigg((x,u);\left(f_{1,\phi}(x,u), f_{2,\phi}(x,u)\right)\bigg)
\end{align}
for some smooth $f_{1,\phi},f_{2,\phi}:\phi[U]\times\Bbb{R}^n\to \Bbb{R}^n$.
Your suggestion for the definition of $\widetilde{F}$ is more explicitly, as follows. You're defining a function $\widetilde{F}_{\phi}:\phi[U]\times\Bbb{R}^n\to (\phi[U]\times\Bbb{R}^n)\times (\Bbb{R}^n\times\Bbb{R}^n)$ as
\begin{align}
\widetilde{F}_{\phi}(x,u):= \bigg((x,u);\left(u, f_{2,\phi}(x,u)\right)\bigg)
\end{align}
If we started with a different chart $(V,\psi)$ on $Q$, we could play the same game. We define $F_{\psi}:= T^2\psi\circ F\circ T\psi^{-1}$ the local representation of the vector field $F$ in terms of the $\psi$ chart, and then we get that $F_{\psi}(y,v)= \bigg((y,v); \left(f_{1,\psi}(y,v),f_{2,\psi}(y,v)\right)\bigg)$ for some smooth $f_{1,\psi},f_{2,\psi}$. Similarly, define $\widetilde{F}_{\psi}(y,v)=\bigg((y,v);(v,f_{2,\psi}(y,v))\bigg)$.
The question then becomes does there actually exist a smooth vector field $\tilde{F}:TQ\to TTQ$ such that for all pairs of charts $\psi$ and $\phi$, we have the (necessary and sufficient) compatibility condition
\begin{align}
\widetilde{F}_{\psi}= \left(T^2\psi\circ T^2\phi^{-1}\right)\circ \widetilde{F}_{\phi}\circ \left(T\psi\circ T\phi^{-1}\right)^{-1}?
\end{align}
In other words, can we piece together the $\widetilde{F}_{\phi}$'s in a consistent manner to get a well-defined smooth vector field $\widetilde{F}:TQ\to TTQ$? Unfortunately the answer is no. By a routine but frustrating (because there's a lot of stuff going on) calculation, you can convince yourself of this negative result, simply by looking at the various transformation laws of the charts. Things just don't match up. (If you really want, maybe later I'll write out explicitly the transitions, but this is a really good exercise in understanding the role of charts, so I strongly recommend you work out these details first).
So, what this means is that there is no "natural operation" relying solely on the smooth structures alone, which takes vector fields $F:TQ\to TTQ$, and produces "Newtonian vector fields" $\widetilde{F}:TQ\to TTQ$.
I can certainly understand the reason why we might wish very strongly for this procedure to work out. After all, we all learn that "velocity is derivative of position, and acceleration is derivative of velocity", so the natural thing to look at is indeed the second tangent bundle. However, we have to be careful about what we mean by "derivative" (note also the remark in @Christoph's answer about $TTQ$ not having the right vector bundle structure to support addition of accelerations). The second tangent bundle, while an interesting space, has a lot of excess information, not suited for modelling accelerations. The acceleration we think of comes from Riemannian geometry and the Levi-Civita connection. Generally, a connection in a vector bundle $(E,\pi,M)$ is what we need in order to differentiate $E$-valued maps without enlarging the target space to $TE$.
@ACuriousMind, and @Christoph have already mentioned the role played by the Ehresmann connection. In this MSE answer, I addressed the issue that we should really think of accelerations as being elements of $TQ$, so perhaps you might find that a helpful read.
