I think you might try approaching this in the Heisenberg picture.
The time derivative of the position operator is:
$$\dfrac{d \hat x}{dt} = \dfrac{i}{\hbar}[\hat H, \hat x]$$
which is a reasonable velocity operator. The time derivative of the velocity operator is then:
$$\dfrac{d^2 \hat x}{dt^2} = \dfrac{i}{\hbar}[\hat H, \dfrac{d \hat x}{dt}]$$
For example, consider a free particle so that $\hat H = \frac{\hat P^2}{2m}$. The velocity operator would then be $\frac{\hat P}{m}$. This certainly looks reasonable as it is of the form of the classical $\vec v = \frac{\vec p}{m}$ relationship.
But, note that the velocity operator commutes with this Hamiltonian so the commutator in the definition of the acceleration operator is 0. But that is what it must be since we're assuming the Hamiltonian of a free particle which means there is no force acting on it.
Now, consider a particle in a potential so that $\hat H = \frac{\hat P^2}{2m} + \hat U$. The velocity operator, for this system, is then $\frac{\hat P}{m} + \frac{i}{\hbar}[\hat U, \hat x]$.
Assuming the potential is not a function of momentum, the commutator is zero and the velocity operator is the same as for the free particle.
The acceleration operator is then $\dfrac{i}{\hbar}[\hat U, \frac{\hat P}{m}]$.
In the position basis, this operator is just $\frac{-\nabla U(\vec x)}{m} $ which looks like the acceleration of a classical particle of mass m in a potential given by $U(\vec x)$.
There is nothing to "derive" for the kinetic energy operator. By definition, classical kinetic energy is $\frac{p^2}{2m}$, and so $\hat{E}_\text{kin} = \frac{\hat{p}^2}{2m}$ quantumly. It's not exactly clear why you think this doesn't make sense mathematically, but it does: In words, it says "apply the momentum operator twice, then divide the result by $2m$".
Note that $\langle p^2\rangle \neq \langle p\rangle^2$, the difference is precisely what the standard deviation is defined as and what is usually called the "uncertainty" $\Delta p$ in most physics texts.
Best Answer
Once a differential operator represents the momentum $p$, $p^2 f$ should be intended as the operator $p$ acting on $pf$. Therefore, the second expression is the correct one.
Edit after some exchange of comments
Probably the key point to eliminate doubts as those raised in the question is that quantum mechanics (QM) and classical mechanics are theories aimed at describing the time evolution of the state of a system and how observable quantities can be evaluated knowing the state.
There are equivalent ways to implement such a description. The Hilbert space representation introduces a dual framework: on the one hand, states are described in terms of the elements of a Hilbert space. On the other hand, observables are represented by operators on the Hilbert space of the states (well, there are some technical subtleties, but I neglect them since I would like to give the overall picture). In the case of usual spinless particles, such representation is based on an algebra of operators generated by the position $q$ and momentum $p$ operators obeying the following commutation rule: $$ [q,p] = i \hbar. $$ All the observables that in classical mechanics are functions of $p$ only are represented by the same function of the operator $p$. Therefore, the kinetic energy is represented by the operator $$ \frac{p^2}{2m}. $$ If we choose the Hilbert space of the square-integrable functions of the position $L^2(\mathbb R^3)$, the representation of the momentum operator is $$ p = -i \hbar \frac{\partial}{\partial x} $$ and the operator representing the kinetic energy must be $$ \frac{p^2}{2m}=- \frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2}. $$