Just to clear up some notational confusion (there doesn't seem to be any completely standard notation):
In Bernstein's notes:
The "easy" pullback (i.e. the one that coincides with the pullback of the underlying $\mathcal O$-module) is denoted $Lf^\Delta$.
$f^! = Lf^\Delta [dim X - dim Y]$ (right adjoint to $f_!$)
$f^\ast = \mathbb D f^! \mathbb D$ (left adjoint to $f_\ast$)
Note that $f^\ast$ and $f^!$ agree with the corresponding functors for constructable sheaves (not for the underlying $\mathcal O$-modules).
However... In Hotta, Takeuchi, Tanisaki
The easy pullback is denoted $Lf^\ast$ to agree with the $\mathcal O$-module functor.
Berstein's $f^!$ is now called $f^\dagger$.
Bernstein's $f^\ast$ is now called $f^\star$
Ok, now in David Ben-Zvi's answer above (and in many other places)
The easy pullback is $f^\dagger$, and the rest agrees with Berstein's notation.
In my opinion, the most important notational feature to be preserved is that $(f^\ast , f_\ast)$ and $(f_! , f^!)$ form adjoint pairs.
To answer your question
When $f$ is smooth, $f^\ast = \mathbb D f^! \mathbb D = f^! [2(dim Y - dim X)]$, and the easy inverse image functor is self dual (and preserves the t-structure).
One way to think about these different pullbacks is that the easy inverse image preserves the structure sheaf $\mathcal O_Y$. This corresponds to the constant sheaf shifted in perverse degree under the RH correspondence. On the other hand $f^\ast$ preserves the "constant sheaf" whereas $f^!$ preserves the dualizing sheaf (for a smooth complex variety, these correspond to the D-modules $\mathcal O[-n]$ and $\mathcal O[n]$).
Let $\mathcal M$ be a left $\mathcal D$-module over a smooth curve $X$. Then the Koszul duality functor assigns to $\mathcal M$ the DG-module $\Omega_X\otimes_{\mathcal O_X}\mathcal M$ over the de Rham complex $\Omega_X$. Viewed as a complex of sheaves, $\Omega_X\otimes_{\mathcal O_X}\mathcal M$ is a two-term complex whose terms are quasi-coherent $\mathcal O_X$-modules, but the differential is not $\mathcal O_X$-linear. For this reason, even when $X$ is affine, one has to distinguish between the acyclicity of the complex of global sections of $\Omega_X\otimes_{\mathcal O_X}\mathcal M$ and the acyclicity of this complex of sheaves itself.
Explicitly, let $x$ be a global coordinate on $X$ (assuming that one exists). Then the differential in the complex $\Omega_X\otimes_{\mathcal O_X}\mathcal M = (\mathcal M\to \Omega_X^1\otimes_{\mathcal O_X}\mathcal M)$ has the form $m\mapsto dx\otimes \partial/\partial x(m)$. The question about acyclicity of this complex of sheaves is, therefore, the question about injectivity and surjectivity of the operator $\partial/\partial x$ acting in the sections of $\mathcal M$.
In the case at hand, we have $\mathcal M=\mathcal O_X e^x$. So global sections of $\mathcal M$ over $\operatorname{Spec} \mathbb C[x]$ are expressions of the form $p(x)e^x$, where $p(x)$ is a polynomial in $x$. Hence one can easily see that the complex of global sections of $\Omega_X\otimes_{\mathcal O_X}\mathcal M$ over $\operatorname{Spec} \mathbb C[x]$ is acyclic.
The complexes of sections of $\operatorname{Spec} \mathbb C[x]$ over Zariski open subsets of $\operatorname{Spec} \mathbb C[x]$ are not acyclic, however. It suffices to consider sections over $\operatorname{Spec} \mathbb C[x,x^{-1}]$. As is well known, the function $x^{-1}e^x$ does not lie in the image of $\partial/\partial x$ acting in the space of Laureant polynomials in $x$ multiplied with $e^x$. Therefore, the two-term complex of sheaves $\Omega_X\otimes_{\mathcal O_X}\mathcal M$ is not acyclic. Its differential is an injective, but not surjective morphism of sheaves of $\mathbb C$-vector spaces over $\operatorname{Spec} \mathbb C[x]$.
On the other hand, one can consider the complex of sheaves of analytic forms $\Omega_X^{an}\otimes_{\mathcal O_X}\mathcal M$ in the analytic topology of the set of closed points of $\operatorname{Spec} \mathbb C[x]$. Then every function from $\mathcal O_X^{an}e^x$ will have a primitive analytic function locally in the analytic topology. So the differential in the two-term complex of sheaves $\Omega_X^{an}\otimes_{\mathcal O_X}\mathcal M$ is now surjective. However, it is no longer injective, as the constant functions are sections of $\mathcal O_X^{an}e^x$, the function $e^{-x}$ being analytic. This is what Kapranov is doing in his paper.
Having, as I hope, answered your question, let me now point out, as a side note, that what I would consider a superior alternative of the $\mathcal D{-}\Omega$ duality theory of the Beilinson--Drinfeld preprint can be found in Appendix B to my AMS Memoir "Two kinds of derived categories, Koszul duality, and comodule-contramodule correspondence", http://arxiv.org/abs/0905.2621 .
The point is that what you call "$\mathcal D$-quasi-isomorphisms" are defined in my paper in terms intrinsic to DG-modules over $\Omega$ (without any reference to differential operators). The corresponding localization is called the coderived category of quasi-coherent DG-modules over $\Omega_X$, and subsequently it is proven to be equivalent to the derived category of quasi-coherent $\mathcal D_X$-modules (for a smooth variety $X$ over any field, $\mathcal D_X$ denoting the crystalline differential operators in the case when the characteristic is finite).
Best Answer
[All functors in this answer are assumed to be derived]
These commutativities basically boil down to unraveling all definitions. The actual computations are quite annoying (but totally possible), so I will only explain the main points.
The diagram commutes for $f^*$ and left $\mathcal{D}$-modules. This basically follows from the definition (do not confuse it with the other $f^*_{\mathcal{D}}$ functor that is defined only on holonomic $\mathcal{D}$-modules).
The functor $f^!$ on left $\cal{D}$-modules is defined as $f^*[\mathrm{dim}_X -\mathrm{dim}_Y]$ for a morphism of smooth varieties $f\colon X \to Y$. In particular $f^!(\mathcal{O}_Y) \simeq \mathcal{O}_X[\mathrm{dim}_X - \mathrm{dim}_Y]$, so there is no chance that the first diagram in your post commutes.
Before we discuss what happens for right $\mathcal{D}$-modules, we need to recall the so-called "left-right switch". This is a canonical equivalence between left and right $\mathcal{D}$-modules on any smooth variety $X$. Namely, it is given by the tensor product functor $-\otimes_{\mathcal{O}_X} \omega_X\colon \mathrm{D}^b_{qc}(\mathrm{LMod}_{\mathcal{D}_X}) \to \mathrm{D}^b_{qc}(\mathrm{RMod}_{\mathcal{D}_X})$. One needs to actually make sense of this, roughly the point is that the canonical bundle $\omega_X$ has a canonical structure of a right $\mathcal{D}$-module via Lie derivatives. Then a tensor product over the structure sheaf of a left $\mathcal{D}$-module and a right $\mathcal{D}$-module has a canonical structure of a right $\mathcal{D}$-module.
Now we are ready to discuss what happens for the $f^!$ functor on the level of right $\mathcal{D}$-modules. In order to avoid any confusion, let us denote it by $f^{\dagger}$. It is defined via the left-right switch. Namely, $f^\dagger(M):= f^!(M\otimes_{\mathcal{O}_Y} \omega^{\vee}_Y)\otimes_{\mathcal{O}_X} \omega_X \colon \mathrm{D}^b_{qc}(\mathrm{RMod}_{\mathcal{D}_Y}) \to \mathrm{D}^b_{qc}(\mathrm{RMod}_{\mathcal{D}_Y})$ for a morphism $f\colon X \to Y$ of smooth varieties. Now we use steps $1$ and $2$ to compute this functor (on the level of underlying $\mathcal{O}$-modules) $$ f^\dagger(M) = f^!(M\otimes_{\mathcal{O}_Y}\omega^{\vee}_Y) \otimes_{\mathcal{O}_X} \omega_X= f^*(M) \otimes_{\mathcal{O}_X} f^*(\omega^\vee_Y)[\mathrm{dim}_X-\mathrm{dim}_Y]\otimes_{\mathcal{O}_X} \omega_X $$ $$ =f^*(M)\otimes_{\mathcal{O}_X} f^*(\omega^{\bullet, \vee}_Y) \otimes_{\mathcal{O}_X} \omega_X^\bullet $$ where $\omega_X^\bullet$ is the dualizing complex on $X$, so $\omega_X^\bullet = \omega_X[\mathrm{dim}_X]$, and the same for $Y$. Now any finite type morphism of regular scheme is of finite Tor-amplitude, so Tag0B6U gives us that $$ f^!(M)=f^*(M)\otimes_{\mathcal{O}_X} f^!(\mathcal{O}_Y) $$ and, essentially by definition, $$ \omega^\bullet_X = f^!(\omega^\bullet_Y) = f^*(\omega^\bullet_Y) \otimes_{\mathcal{O}_X} f^!(\mathcal{O}_Y). $$ Combining these two equalities, we get the desired formula $$ f^!(M) = f^*(M)\otimes_{\mathcal{O}_X} \omega^\bullet_X \otimes_{\mathcal{O}_X} f^*(\omega^{\bullet, \vee}_Y) = f^\dagger(M) $$
The diagram for $f_*$ does not commute neither for left nor for right $\mathcal{D}$-modules because it is neither left nor right $t$-exact on the $\mathcal{D}$-module side (consider separately cases of a closed immersion and a smooth morphism) but it is left exact on the $\mathcal{O}$-module side. However, one can ask for commutativity of another diagram that relates $\mathcal{O}$-modules and $\mathcal{D}$-modules. Namely, given any object $K\in \mathrm{D}_{qc}^b(\mathcal{O}_Y)$, one can define an induced $\mathcal{D}$-module $K\otimes_{\mathcal{O}_Y}\mathcal{D}_Y \in \mathrm{D}^b_{qc}(\mathcal{D}_Y)$. Then the claim is that for the right $\mathcal{D}$-modules, there is a functorial isomorphism $$ f_*(K\otimes_{\mathcal{O}_X} \mathcal{D}_X) \simeq f_*(K) \otimes_{\mathcal{O}_Y} \mathcal{D}_Y $$ for a morphism $f\colon X \to Y$ of smooth varieties. The proof essentially boils down to the projection formula (and a bit tedious computations since you want to identify those objects as $\mathcal{D}$-modules and not merely $\mathcal{O}$-modules) once all definitions are given (but it is quite tricky to define $f_*$ on the $\mathcal{D}$-module side). A version of this result is Lemma 4.26 in Kashiwara's book "$\mathcal{D}$-modules and microlocal analysis" (actually formulated for left $\mathcal{D}$-modules).
Commutativity of your last diagram depends on your definition of duality on both sides, I think.