2 Minor edit for clarity.

This is not a helpful answer to your main question, but merely a negative answer to your "In particular..what happens.." question. But, the general idea may be helpful in figuring out what more precise things (weaker than Keller's result) would be reasonable to ask for.

$\newcommand{\RR}{\mathbb{R}}\newcommand{\RHom}{\mathrm{RHom}}\newcommand{\pt}{\mathrm{pt}}\DeclareMathOperator{\deg}{deg}\newcommand{\RGamma}{R\Gamma}\renewcommand{\mod}{\text{-mod}}\newcommand{\C}[1]{C^\bullet(#1)}$

Since I'll lapse into this notation anyway, let me make it explicit: Identity $\gamma_M$ with the functor $$\RGamma(M, -)\colon D \to \C{M}\mod$$ where by $\mod$ I'm implicitly working in a dg-setting.

At some point below, I'll assume that $N$ is compact oriented of dimensions $n$. (I guess you had it simply connected anyway...This is not strictly necessary, but allows me to avoid extra notation.)

Claim: Suppose that $M$ is simply connected, $\dim M \geq 2$, and that $i \colon N \hookrightarrow M$ is not the identity. Then, the functor $\gamma_M \colon D D= \langle \RR_M, i_* \RR_N \rangle \to D_{A_M}$ is not fully faithful.

## Fuzzy Remark:

Before sketching an argument, here's a "philsophical" remark about why things will go wrong:

The category $D_c(M)$ feels the topology (or maybe even geometry) of $M$. In particular, it has a Proper Base-change Theorem saying something like $q^* p_! = (p')_! (q')^*$ where the maps take part in a fiber-square of (actual) topological spaces. The category $D_{A_M} = \C{M}\mod$ feels only the homotopy theory of $M$. You should expect a Base-Change Theorem in this context, but now with a fiber-square of homotopy types -- more correctly, a homotopy fiber-square.

A simpler sort of 'no go' result that this heuristic implies: Suppose you had wanted to include two sub-manifolds $i_k\colon N_k \hookrightarrow M$, $k=1,2$. The $\C{M}\mod$ images would be unable to tell them apart if the $i_k$ were homotopic -- e.g., the inclusion of any two points. While the constructible theory would certain care whether the two points were the same or not.

## Sketch of Claim:

To see this, note that $$\RHom_{D_c(M)}(i_* \RR_N, i_* \RR_N) = \RHom_{D_c(N)}(i^* i_* \RR_N, \RR_N) = \RHom_{D_c(N)}(\RR_N, \RR_N) = \C{N}$$ while

Sub-Claim: Letting $\stackrel{h}\times_M$ denote the homotopy fiber product, $$\RHom_{A_M}(\gamma_M(i_* \RR_N), \gamma_M(i_* \RR_N)) =\RHom_{\C{M}}\left(\C{N}, \C{N}\right) \approx C_{\bullet}\left(N \stackrel{h}\times_M N\right)[-n]$$

Assuming the sub-claim: to conclude it suffices to produce homology classes on $\Omega M$ in arbitrarily positive degrees, whose images under the composite $$H(\Omega M) \to H_*(N \stackrel{h}\times_M N) \to H_*(\Omega (M/N))$$ are non-zero. I think the following should do this upon filling in the details: Equip $M$ with a base-point in $N$, take some non-zero element of $\pi_i M$, with $i \geq 2$, that remains non-zero in $\pi_i (M/N)$. Use it to produce an $(i-1)$-homology class on $\Omega M$, and then take its Pontrjagin products.

Sketch of sub-claim: Underlying the Eilenberg-Moore spectral sequence is the statement that, letting $\boxtimes$ denote derived co-tensor of co-modules over a co-algebra, $$C_\bullet(N) \boxtimes_{C_\bullet(M)} C_\bullet(N) \approx C_\bullet(N \stackrel{h}\times_M N)$$ Poincare duality gives an equivalence $\C{N} \approx C_\bullet(N)[-n]$ of $\C{M}$-modules (or $C_\bullet(M)$-comodules). It remains to identify $$\RHom_{\C{M}}(\C{N}, \C{N}) \approx C_\bullet(N) \boxtimes_{C_\bullet(M)} \C{N}$$ by term-wise identifying the co-simplicial cobar constructions on both sides.

Example: Note that if $N = \pt$ the sub-claim is a familiar statement in Koszul duality: That for $M$ simply-connected $\RHom_{\C{M}}(\RR,\RR) \approx C_\bullet(\Omega M)$. In certain cases, e.g. $M = S^{2k+1}$, you can just see it. As an aside: $C_\bullet(\Omega M)\mod$ knows about all locally-constant things, not just local systems finitely-buildable from the trivial one. (But will run into the same issues if you try to include submanifolds without explicitly adding in extra generators for the strata.)

Remark: Though $\gamma_M$ is not fully-faithful here, it does get the maps into/out of $\RR_M$ right. Logic as above shows that $$\RHom_D(\RR_M, i_* \RR_N) = C^*(\RR_N) \C{N} = \RHom_{\C{M}}(\C{M}, \C{N})$$ and then Verdier Duality (for $D_c(M)$) + something like Grothendieck Duality (for $\C{M}$) give the other direction as well.

1

This is not a helpful answer to your main question, but merely a negative answer to your "In particular..what happens.." question. But, the general idea may be helpful in figuring out what more precise things (weaker than Keller's result) would be reasonable to ask for.

$\newcommand{\RR}{\mathbb{R}}\newcommand{\RHom}{\mathrm{RHom}}\newcommand{\pt}{\mathrm{pt}}\DeclareMathOperator{\deg}{deg}\newcommand{\RGamma}{R\Gamma}\renewcommand{\mod}{\text{-mod}}\newcommand{\C}[1]{C^\bullet(#1)}$

Since I'll lapse into this notation anyway, let me make it explicit: Identity $\gamma_M$ with the functor $$\RGamma(M, -)\colon D \to \C{M}\mod$$ where by $\mod$ I'm implicitly working in a dg-setting.

At some point below, I'll assume that $N$ is oriented of dimensions $n$. (I guess you had it simply connected anyway...)

Claim: Suppose that $M$ is simply connected, $\dim M \geq 2$, and that $i \colon N \hookrightarrow M$ is not the identity. Then, the functor $\gamma_M \colon D \to D_{A_M}$ is not fully faithful.

## Fuzzy Remark:

Before sketching an argument, here's a "philsophical" remark about why things will go wrong:

The category $D_c(M)$ feels the topology (or maybe even geometry) of $M$. In particular, it has a Proper Base-change Theorem saying something like $q^* p_! = (p')_! (q')^*$ where the maps take part in a fiber-square of (actual) topological spaces. The category $D_{A_M} = \C{M}\mod$ feels only the homotopy theory of $M$. You should expect a Base-Change Theorem in this context, but now with a fiber-square of homotopy types -- more correctly, a homotopy fiber-square.

A simpler sort of 'no go' result that this heuristic implies: Suppose you had wanted to include two sub-manifolds $i_k\colon N_k \hookrightarrow M$, $k=1,2$. The $\C{M}\mod$ images would be unable to tell them apart if the $i_k$ were homotopic -- e.g., the inclusion of any two points. While the constructible theory would certain care whether the two points were the same or not.

## Sketch of Claim:

To see this, note that $$\RHom_{D_c(M)}(i_* \RR_N, i_* \RR_N) = \RHom_{D_c(N)}(i^* i_* \RR_N, \RR_N) = \RHom_{D_c(N)}(\RR_N, \RR_N) = \C{N}$$ while

Sub-Claim: Letting $\stackrel{h}\times_M$ denote the homotopy fiber product, $$\RHom_{A_M}(\gamma_M(i_* \RR_N), \gamma_M(i_* \RR_N)) =\RHom_{\C{M}}\left(\C{N}, \C{N}\right) \approx C_{\bullet}\left(N \stackrel{h}\times_M N\right)[-n]$$

Assuming the sub-claim: to conclude it suffices to produce homology classes on $\Omega M$ in arbitrarily positive degrees, whose images under the composite $$H(\Omega M) \to H_*(N \stackrel{h}\times_M N) \to H_*(\Omega (M/N))$$ are non-zero. I think the following should do this upon filling in the details: Equip $M$ with a base-point in $N$, take some non-zero element of $\pi_i M$, with $i \geq 2$, that remains non-zero in $\pi_i (M/N)$. Use it to produce an $(i-1)$-homology class on $\Omega M$, and then take its Pontrjagin products.

Sketch of sub-claim: Underlying the Eilenberg-Moore spectral sequence is the statement that, letting $\boxtimes$ denote derived co-tensor of co-modules over a co-algebra, $$C_\bullet(N) \boxtimes_{C_\bullet(M)} C_\bullet(N) \approx C_\bullet(N \stackrel{h}\times_M N)$$ Poincare duality gives an equivalence $\C{N} \approx C_\bullet(N)[-n]$ of $\C{M}$-modules (or $C_\bullet(M)$-comodules). It remains to identify $$\RHom_{\C{M}}(\C{N}, \C{N}) \approx C_\bullet(N) \boxtimes_{C_\bullet(M)} \C{N}$$ by term-wise identifying the co-simplicial cobar constructions on both sides.

Example: Note that if $N = \pt$ the sub-claim is a familiar statement in Koszul duality: That for $M$ simply-connected $\RHom_{\C{M}}(\RR,\RR) \approx C_\bullet(\Omega M)$. In certain cases, e.g. $M = S^{2k+1}$, you can just see it. As an aside: $C_\bullet(\Omega M)\mod$ knows about all locally-constant things, not just local systems finitely-buildable from the trivial one. (But will run into the same issues if you try to include submanifolds without explicitly adding in extra generators for the strata.)

Remark: Though $\gamma_M$ is not fully-faithful here, it does get the maps into/out of $\RR_M$ right. Logic as above shows that $$\RHom_D(\RR_M, i_* \RR_N) = C^*(\RR_N) = \RHom_{\C{M}}(\C{M}, \C{N})$$ and then Verdier Duality (for $D_c(M)$) + something like Grothendieck Duality (for $\C{M}$) give the other direction as well.