The piston which is referenced in basic thermodynamics is not a real, dynamical object with mass or inertia. It's an idealization that serves only as a "handle" by which the volume of the system can be changed.
To understand the model, consider an ideal gas with a fixed $N$. The state of the system is a point in $(P,V,T)$ space which lies on a surface
$$\frac{PV}{T} = const$$
A quasi-static thermodynamic process is a curve embedded in this surface. In particular, an isothermal process is a curve restricted to constant $T$, shown here:
From a mathematical perspective, that's all there is to it - we fix $T$ to be constant and observe that $P$ and $V$ must evolve such that $PV=const$. However, to make the connection with a physical system (e.g. a gas), we talk about how we would manipulate the system into behaving this way.
When we restrict the system to a constant $T$, we say we place it in contact with a thermal reservoir. When we restrict it to constant $V$, we say we lock the piston. When we restrict it to constant $P$, we say that we expose the piston to a fixed external pressure. However, when we do so we are not modeling the physical mechanisms by which these constraints are being imposed. "Heat bath" and "piston" and "external pressure" are just the words that we drape over the idea that the temperature, volume, and pressure of our gas can be constrained or manipulated at will.
That's not to say that we can't model these things. You can give the piston some real mass $m$ and introduce frictional forces on it. However, if you do this then you're going to have to couple Newton's 2nd law (applied to the piston) to the laws of thermodynamics (applied to the gas). You would also have to assume that any dynamics occur on a time scale sufficiently slow to make thermodynamics applicable in the first place.
The initial work, $f\ dx = \delta Q$, at the very beginning will give the piston momentum [...]
Unless you are genuinely modeling the piston as a real physical object, you should not go down this road. In real life, the work done by the gas is done on the external environment as well as on the piston. In idealized thermodynamics, we neglect the latter because the piston is not to be thought of as a real object.
In the spirit of demonstrating my point, consider a simple case of isobaric expansion of an ideal gas. We will consider the process to be a sequence of adiabatic steps, during which the pressure increases and then relaxes to the ambient pressure.
Let the external pressure on the system be $P_{ext}=P_0-mg/A$ (the gas must be at a slightly higher pressure than the environment to support the weight of the piston), and the initial pressure, temperature, and volume of the gas be $P_0,T_0,$ and $V_0$. The system begins at equilibrium, so $T_0 = P_0 V_0/N$ (I choose units in which $k_B=1$ for convenience).
The piston has mass $m$, cross-sectional area $A$, and, when moving, experiences a frictional force $\vec F=-2m\Gamma \vec v$ due to its contact with the walls. Its initial position is $y=y_0 = V_0/A$.
At the time $t=0$, a small amount of heat $\delta Q$ is delivered to the gas. We assume that this energy is distributed instantly and uniformly throughout the system (at least, before the volume can change), thereby raising its temperature by an amount $\delta T = \frac{2\delta Q}{3N}$. Since the volume does not change, this raises the pressure by an amount $\delta P = \frac{N}{V_0} \delta T = \frac{2\delta Q}{3 V_0}$. No more heat is added to the system.
We now want to know the details of the systems "relaxation" (or equilibriation). The equations which govern the evolution of the system are Newton's 2nd law and the ideal gas law. They are:
$$m \ddot y(t) = A(P-P_{ext}) - mg - m\Gamma \dot y(t) $$
$$PV = NT$$
with the latter constrained by the adiabatic condition
$$PV^\gamma = const$$
with $\gamma = 5/3$.
Newton's law can be rearranged to yield
$$\ddot y(t) + 2\Gamma \dot y(t) = \frac{A}{m}(P-P_0)$$
while the adiabatic condition yields
$$PA^\gamma y^\gamma = (P_0+\delta P)A^\gamma y_0 \implies P = (P_0+\delta P) \left(\frac{y_0}{y}\right)^\gamma$$
Therefore, we have
$$\ddot y(t) + 2\Gamma \dot y(t) - \frac{P_0 A}{m}\left(\left(1+\frac{\delta P}{P_0}\right)\left[\frac{y_0}{y}\right]^\gamma-1\right) = 0 $$
In general this is not an easy problem to solve. However, recall that the amount of heat added (and therefore the change in volume) is very small. It follows that we would expect $y=y_0 + z(t)$ where $z(t)\ll y_0$. In that approximation,
$$\left(\frac{y_0}{y_0 + z}\right)^\gamma \approx \left(1-\frac{z}{y_0}\right)^\gamma \approx 1 - \gamma\frac{z}{y_0}$$
$$\left(1+\frac{\delta P}{P_0}\right)\left(1-\gamma \frac{z}{y_0}\right) \approx 1 - \gamma \frac{z}{y_0} + \frac{\delta P}{P_0}$$
and so our equation becomes
$$\ddot z(t) + 2\Gamma \dot z(t) + \omega^2 z = \frac{A\delta P}{m}$$
This is the equation of the damped harmonic oscillator under a constant external force, where we've defined $\omega^2 \equiv \frac{P_0 A \gamma}{m y_0}$$ for convenience.
Our initial condition corresponds to $z(0)=0$ and $\dot z(0) = 0$. The homogeneous solution to the equation is
$$z_H(t) = e^{-\Gamma t}\left(c_1 e^{\alpha t} + c_2 e^{-\alpha t}\right)$$
where $\alpha \equiv \sqrt{\Gamma^2-\omega^2}$. The inhomogeneous solution is simply a constant $z_\infty$ - plugging it into the equation yields $\omega^2 z_\infty = \frac{\delta P}{P_0} \implies z_\infty = \frac{A\delta P}{\omega^2 m}$.
Imposing the initial conditions $z(0) = \dot z(0) = 0$, we find
$$z(t) = z_\infty \left( 1 - e^{-\Gamma t}\left[\frac{\Gamma}{\alpha}\sinh(\alpha t) + \cosh(\alpha t)\right]\right)$$
where $z_\infty \equiv \frac{A\delta P}{\omega^2 m} = \frac{2 \delta Q}{3\gamma P_0 A}$.
Now let's examine this result. The change in volume is $\delta V = z_\infty A = \frac{2\delta Q}{3 \gamma P_0}$. The final pressure is simply $P_0$, and the final temperature is $T = T_0 + \delta T = T_0 + \frac{2 \delta Q}{3 \gamma N}$. Assuming that the damping is sufficient to prevent oscillatory behavior (i.e. $\Gamma^2 \gg \omega^2$), the time scale for this expansion is
$$\tau^{-1} \approx \Gamma - \sqrt{\Gamma^2-\omega^2} \approx \frac{\omega^2}{2\Gamma}$$
$$\implies \tau = \frac{2\Gamma}{\omega^2} = \frac{2m \Gamma y_0}{P_0 A \gamma}$$
If we take $M$ such steps, then we would find $M$ times the total change in temperature and volume (i.e. $\Delta V = M \cdot \delta V$, $\Delta T = M \cdot \delta T$, $\Delta Q = M \cdot \delta Q$).
Compare this with the vastly simpler thermodynamics problem. If we supply a total quantity of heat $\Delta Q$ to an ideal gas at constant pressure, then its temperature will increase according to $\Delta Q = c_P \Delta T = \frac{5}{2}N\Delta T \rightarrow \Delta T = \frac{2\Delta Q}{5N}$. Since $\Delta U = \frac{3}{2}N\Delta T$ and $\Delta U = \Delta Q - P \Delta V$, it follows that
$$\Delta V = \frac{1}{P}(\Delta Q - \Delta U) = \frac{1}{P}\left(\Delta Q - \frac{3}{2}N \frac{2}{5N} \Delta Q\right) = \frac{2 \Delta Q}{5P}$$
and
$$\Delta T = \frac{P}{N} \Delta V = \frac{2 \Delta Q}{5N}$$
