A quick follow-on to this question. Consider the following SDE: $$ \ddot{x} = f(x) - \gamma g(x)\, \dot{x} + \sigma h(x)\, \xi(t) \tag{1} $$
Based on [1], we can represent (1) as a system of two first-order equations: $$ \begin{cases} \mathrm{d} X_t = V_t\, \mathrm{d}t \\ \mathrm{d} V_t = -\gamma g(X_t)\, \mathrm{d}t + f(X_t)\, \mathrm{d}t + \sigma h(X_t)\, \mathrm{d}W_t. \end{cases} \tag{2} $$
Why does the noise contribute to the $V_t$ variables, not the $X_t$ variables? Is it something particular about (1)? Can we rewrite (2) equivalently with noise on the $X_t$ variables?
[1] Burrage, Kevin; Lenane, Ian; Lythe, Grant, Numerical methods for second-order stochastic differential equations, SIAM J. Sci. Comput. 29, No. 1, 245-264 (2007). ZBL1144.65004.
It really depends on the physics hidden behind, namely on where you think the stochastics finds its role. Let me explain with two examples. In both examples, $X_t$ denotes the position vector, $V_t$ the "velocity", and $W_t$ the Wiener process.
The first example is the Brownian motion. If we consider the standard Brownian motion, its equation of motion is obviously $$ {\rm d}X_t={\rm d}W_t. $$ Here we believe, in physics, that the stochastics contributes to the position directly. More precisely, we owe the physically-observed stochastic behavior of tiny particles to some mathematically-defined stochastic processes. In this example, there is no properly defined velocity. In other words, if you want to put $$ {\rm d}X_t=V_t\,{\rm d}t+{\rm d}W_t, $$ it is a must that $V_t=0$, meaning that the velocity is zero. This is of course counterintuitive; after all, all Brownian particles are moving, right? So this really depends on how you define velocity for a stochastic process, or an Ito process for special. Mathematically, there are two ways. The first is to identify the drift term as the velocity, i.e., if $$ {\rm d}X_t=V_t\,{\rm d}t+\sigma_t\,{\rm d}W_t, $$ $V_t$ is then the velocity. However, this does not apply to the Brownian motion. The second is to identify the quadratic variation as the velocity, i.e., following the above SDE, $$ {\rm d}\left<X\right>_t=\sigma_t^2\,{\rm d}t, $$ and $\sigma_t$ is then regarded as the velocity. This applies to the Brownian motion, and $\left<X\right>_t$ measures the average kinetic energy.
The second example is your Langevin equation. In this scenario, recall that a Langevin equation models a very large particle whos dynamics is governed by Newton's second law, but with stochastic background forces contributed from tiny Brownian particles. With this understanding, the equation of motion for this very large particle should be \begin{align} {\rm d}X_t&=V_t\,{\rm d}t,\\ {\rm d}V_t&=I_{t+{\rm d}t}-I_t={\rm d}I_t, \end{align} which is in the form of Newton's second law. Here $I_t$ denotes the impulse of external forces. The only difference is that $I_t$ is no longer deterministic, but yields some stochastic behaviors. Therefore, we put $$ {\rm d}I_t=F_t\,{\rm d}t+\sigma_t\,{\rm d}W_t, $$ where $F_t$ stands for the conventional forces, and $\sigma_t$ indicates the strength of stochastic forces. These stochastic forces arise from the Brownian motion of those tiny particles. Combine the last three equations, and you recover the Langevin equation. In this case, the stochastic factors are all put on the forcing term.
Of course, you may consider \begin{align} {\rm d}X_t&=V_t\,{\rm d}t+\mu_t\,{\rm d}W_t^1,\\ {\rm d}V_t&=F_t\,{\rm d}t+\sigma_t\,{\rm d}W_t^2, \end{align} where $W_t^1$ and $W_t^2$ are two Wiener processes. This model is no longer Langevin, but for some intermediate particle instead. In other words, the dynamics of this particle is partly governed by Newton's second law with stochastic background forces, but itself also observes some Brownian motion. It really depends on how you interpret the physics hidden behind.
That's it. Hope this could be somewhat helpful for you.