TL;DR: In general, this cannot be done exactly, because of the relationship between the eigenvectors of $A$ and $e^A$.

That said, the Feynman's clock construction gets pretty close to realizing the desired Hamiltonian. In fact, if post-selection is available, the construction provides an exact realization. Otherwise, Feynman's clock with a ballistic program counter allows us to make the probability of successful realization of the desired Hamiltonian as high as we wish. Perhaps this sheds some light on why the physical intuition expressed in the question works in classical mechanics, but fails in quantum mechanics.

Impossibility of exact solution

First choose the auxiliary subsystems so that $V$, $V_1$ and $V_2$ are all unitaries on the same Hilbert space. This is possible since every isometry can be extended to a unitary. Next, choose two orthonormal bases $|\psi_k\rangle$ and $|\phi_k\rangle$ whose elements are pairwise distinct under the global phase equivalence, i.e. $|\langle\psi_i|\phi_j\rangle|<1$ for all $i,j$. We construct our counterexample as follows $$ V_1=\sum_k|\phi_k\rangle\langle\psi_k|,\quad V_2=\sum_k e^{-ia_k}|\psi_k\rangle\langle\phi_k|\tag1 $$ where $a_k\in\mathbb{R}$ are chosen so that $e^{ia_k}$ are distinct. From $V=V_2V_1$, we have $$ V=\sum_k e^{-ia_k}|\psi_k\rangle\langle\psi_k|\tag2 $$ which means that $|\psi_k\rangle$ are the normalized eigenvectors of $V$.

Now, if $|v\rangle$ is eigenvector of operator $A$ associated with eigenvalue $\lambda$ then $|v\rangle$ is an eigenvector of $e^A=\sum_{k=0}^\infty\frac{A^k} {k!}$ associated with eigenvalue $e^\lambda$. In general the converse may not be true: there may exist $|u\rangle$ that is an eigenvector of $e^A$ but not an eigenvector of $A$. This happens when $|u\rangle$ is a linear combination of eigenvectors of $A$ associated with distinct eigenvalues of $A$ that are mapped by the exponential function to the same eigenvalue of $e^A$. In this case, $e^A$ has repeated eigenvalues.

However, $V$ has distinct eigenvalues, so if $H'$ is any Hamiltonian such that $V=e^{-iH't/\hbar}$ then $|v\rangle$ is an eigenvector of $V$ if and only if it is an eigenvector of $H'$. Consequently, all invariant subspaces of $H'$ are one-dimensional and are spanned by the vectors $|\psi_k\rangle$. But, by the choice of $|\phi_k\rangle$, the vectors $|\psi_k\rangle$ are not eigenvectors of $V_1$ and we conclude that $V_1\ne e^{-iH't_1/\hbar}$.

Feynman's clock and post-selection

We are now going to do what classical CPU does: use a program counter to orchestrate orderly execution of a sequence of discrete instructions. The program counter is an extra auxiliary system $\mathcal{C}$ that we adjoin to $\mathcal{H}\otimes\mathcal{K}$ and the Feynman's clock Hamiltonian acts on it using creation and annihilation operators to execute discrete instructions within a continuous time evolution.

We will think of $\mathcal{C}$ as a particle hopping between sites in a one-dimensional lattice (for now we only need three sites of the lattice). Let $c_i^\dagger$ and $c_i$ denote the creation and annihilation operators associated with site $i$. Let $H_1$ and $H_2$ denote some Hamiltonians acting on $\mathcal{H}\otimes\mathcal{K}$ that generate $V_1$ and $V_2$ $$ V_1=e^{-iH_1t_1/\hbar},\quad V_2=e^{-iH_2t_2/\hbar}.\tag3 $$ Also, define interaction strengths $g_1=\hbar/t_1$ and $g_2=\hbar/t_2$. Then, the Feynman's clock Hamiltonian $H'$ acting on $\mathcal{H}\otimes\mathcal{K}\otimes\mathcal{C}$ is $$ H'=g_1H_1c_1^\dagger c_0+g_2H_2c_2^\dagger c_1+\text{h.c.}\tag4 $$ Note that $H'$ conserves the number of particles in the program counter. We assume there is exactly one such particle and that at $t=0$ it is located in site $0$. If at time $t=t_1$ we find the particle in site $1$ then $H'$ will have applied $V_1$ to $\mathcal{H}\otimes\mathcal{K}$ subsystem. Similarly, if at $t=t_1+t_2$ we observe the particle in site $2$ then $H'$ will have applied $V_2V_1$ to $\mathcal{H}\otimes\mathcal{K}$ subsystem. Note that the hermitian conjugate terms in $(4)$ hide the fact that the particle might propagate in both directions. Thus, if at $t=t_1$ or $t=t_1+t_2$ we observe the particle back in site $0$ then the unitaries on $\mathcal{H}\otimes\mathcal{K}$ will have been undone. This is why post-selection is needed: the system has applied $V_2V_1$ to $\mathcal{H}\otimes\mathcal{K}$ only if we observe the program counter particle in site $2$.

Ballistic program counter

We can boost the probability of that happening arbitrarily close to one by turning the program counter into a ballistic particle. To that end, we extend the lattice in both directions$^1$ and create a wave packet propagating from the sites with negative indices in the positive direction. After sufficient time has passed the probability of the program counter particle to be in a site $i$ with $i\geqslant 2$ will be arbitrarily close to one. Thus, the probability that our system will have aplied $V_2V_1$ to $\mathcal{H}\otimes\mathcal{K}$ can be made as close to one as we wish. We cannot achieve deterministic success this way since it would require a wave packet with compact support in both position and momentum space.

I think this construction can be seen as realizing the physical intuition expressed in the question.

^{$^1$ We also need to add appropriate hopping terms to $H'$.}