This question is possibly related to this other one.

Let $\mathcal A$ be a dg-category over a commutative ring $k$. I denote by $\text{dgm-}\mathcal A$ the dg-category of right dg-$\mathcal A$-modules, that is, dg-functors \begin{equation} \mathcal A^{\text{op}} \to \mathbf C_{\text{dg}}(k), \end{equation} where $\mathbf C_{\text{dg}}(k)$ is the dg-category of cochain complexes of $k$-modules. In general, the category of dg-functors $\mathrm{Fun}_{\text{dg}}(\mathcal A, \mathcal B)$ between two dg-categories is itself a dg-category, with hom-complexes denoted by $\mathrm{Nat}_{\text{dg}}(F,G)$ (the complex of natural transformations $F \to G$).

Let me recall the (differential graded) Yoneda lemma. Given $F \in \text{dgm-}\mathcal A$, we have a natural isomorphism of complexes: \begin{equation} \mathrm{Nat}_{\text{dg}}(\mathcal A(-,A),F) \cong F(A). \end{equation} We also have the differential graded Yoneda embedding: \begin{align} h_{-} : \mathcal A \hookrightarrow \text{dgm-}\mathcal A, \\ A \mapsto \mathcal A(-,A). \end{align}

A useful operation on dg-modules is taking cohomology. In fact, there is a functor \begin{equation} H^0(-) : H^0(\text{dgm-}\mathcal A) \to \text{mod-}H^0(\mathcal A), \end{equation} which takes a dg-module $F$ to the $H^0(\mathcal A)$-module of its zeroth cohomology.

Now, notice that the zeroth cohomology of a representable module $\mathcal A(-,A)$ is given by $H^0(\mathcal A(-,A)) = H^0(\mathcal A)(-,A)$. So, this $H^0(\mathcal A)$-module is also represented by $A$. Then, we can apply both the differential graded and the ordinary version of Yoneda lemma: \begin{align} & H^0(\mathrm{Nat}_{\text{dg}}(\mathcal A(-,A),F)) \cong H^0(F(A)) \\ &= H^0(F)(A) \cong \mathrm{Nat}(H^0(\mathcal A(-,A)), H^0(F)), \end{align} and it is immediate to see that this chain of maps gives actually the functorial map \begin{equation} H^0(\mathrm{Nat}_{\text{dg}}(\mathcal A(-,A),F)) \to \mathrm{Nat}(H^0(\mathcal A(-,A)), H^0(F)) \end{equation} given by the above $H^0(-)$ functor. So, we have proven that in this case, this natural map is an isomorphism.

Now, let me take another step. I start from a closed degree zero map $f: A \to B$ in $\mathcal A$, which corresponds to a closed degree zero morphism $h_f : h_A \to h_B$ via the Yoneda embedding. The category $\text{dgm-}\mathcal A$ is (strongly) pretriangulated, so we can take the cone $C(h_f)$, which fits in the following pretriangle: \begin{equation} h_A \xrightarrow{h_f} h_B \to C(h_f) \to h_A[1]. \end{equation} Now, let $F \in \text{dgm-}\mathcal A$, and consider the dg-functor \begin{equation} \mathrm{Nat}_{\text{dg}}(-, F) : (\text{dgm-}\mathcal A)^{\text{op}} \to \mathbf{C}_{\text{dg}}(k). \end{equation} As a dg-functor, it maps pretriangles to pretriangles (caveat: because of contravariance, the sign of the shift is changed to the opposite). Hence, we obtain a pretriangle in $\mathbf{C}_{\text{dg}}(k)$: \begin{equation} \mathrm{Nat}_{\text{dg}}(h_B,F) \to \mathrm{Nat}_{\text{dg}}(h_A,F) \to \mathrm{Nat}_{\text{dg}}(C(h_f)[-1],F) \to \mathrm{Nat}_{\text{dg}}(h_B[-1],F). \end{equation} Moreover, Yoneda lemma gives a commutative square:

so we obtain a (functorial and strict!) isomorphism between cones:
\begin{equation}
\mathrm{Nat}_{\text{dg}}(C(h_f)[-1],F) \cong C(F(f))
\end{equation}
as cochain complexes. Rearranging shifts, we obtain:
\begin{equation}
\mathrm{Nat}_{\text{dg}}(C(h_f),F) \cong C(F(f))[-1].
\end{equation}
Now, I'd like to compare $H^0(\mathrm{Nat}_{\text{dg}}(C(h_f),F)) \cong H^0(C(F(f))[-1])$ with $\mathrm{Nat}(H^0(C(h_f)),H^0(F))$. Is it true or false that the natural map
\begin{equation*}
H^0(\mathrm{Nat}_{\text{dg}}(C(h_f),F)) \to \mathrm{Nat}(H^0(C(h_f)),H^0(F))
\end{equation*}
is an isomorphism? I expect the answer to be *no*, because, roughly speaking, taking cohomology "destroys" the functoriality of cones. How can I find a counterexample?