The above answer is very good, but some future readers have been confused. Here is the same in more detail, because I am procrastinating :)
There are various possibly different(?) choices of isomorphism $\check{H^\bullet}(\mathfrak{U},M;\Bbb R)\cong H^\bullet_{dR}(M)$ you could make, for a good covering; your linked post suggest you have the following one in mind:
With Koszul sign conventions, make the first quadrant double complex $T=\mathfrak{C}^\bullet(\mathfrak{U},M;\Omega_M^\bullet)$ by applying the functor $\mathfrak{C}^\bullet(\mathfrak{U},M;-):\mathrm{Sh}(M)\to\mathrm{Ch}_{\ge0}(\mathsf{Ab})$ to the de Rham sheaves. There are homomorphisms of double complexes $\phi:\Omega_M^\bullet\to T,\psi:\mathfrak{C}^\bullet(\mathfrak{U},M;\Bbb R)\to T$ where the single complexes are being viewed as double complexes concentrated in one row, one column respectively (the order doesn't matter, so long as you make a consistent choice once and for all; for sign reasons, we would have to replace - say - $\Omega_M^\bullet$ with the complex you achieve by alternating the signs of its differentials).
For spectral sequence (or an easier special case) and good cover reasons, $\phi,\psi$ are both quasiisomorphisms and we are done.
To check this isomorphism is compatible with the Cech restriction maps, all you need to understand is that $\phi,\psi$ are 'stable' or 'natural' with respect to the restrictions; $T$ and $\mathfrak{C}^\bullet(\mathfrak{U},M;-)$ are indeed semi-functorial in the cover $\mathfrak{U}$ (exactly functorial once you take cohomology or if you fixed choices of refinement data; the restriction maps aren't well-defined on the nose, but they are on cohomology classes). In the case of $\psi$, which is just $\mathfrak{C}^\bullet(\mathfrak{U},M;i)$ for $i:\underline{\Bbb R}\hookrightarrow\Omega_M^\bullet$, a fixed refinement $\alpha:\mathfrak{V}\preceq\mathfrak{U}$ gives a natural transformation $\mathfrak{C}^\bullet(\mathfrak{U},M;-)\to\mathfrak{C}^\bullet(\mathfrak{V},M;-)$ and we are happy. Explicitly, "we are happy" means that $\check{H^\bullet}(\mathfrak{U},M;\Bbb R)\to H^\bullet(T_{\mathfrak{U}})\to H^\bullet(T_{\mathfrak{V}})$ equals $\check{H^\bullet}(\mathfrak{U},M;\Bbb R)\to\check{H^\bullet}(\mathfrak{V},M;\Bbb R)\to H^\bullet(T_{\mathfrak{V}})$, because these guys are equal at the level of (double) complexes. If you don't believe me, just think about it; the restriction map I'm using for the double complex $T$ is just the usual restriction maps for the Cech complexes, just we have to apply it to all the de Rham sheaves in the "rectangular array" and observe that yes, really, that gives a homomorphism of double complexes. But since the restriction operation is exactly the same, and you should know more simply that: $$\require{AMScd}\begin{CD}\mathfrak{C}^n(\mathfrak{U},M;\mathscr{F})@>>>\mathfrak{C}^n(\mathfrak{U},M;\mathscr{G})\\@VVV@VVV\\C^n(\mathfrak{V},M;\mathscr{F})@>>>C^n(\mathfrak{V},M;\mathscr{G})\end{CD}$$Commutes for any cover-refinement choice and homomorphism of sheaves $\mathscr{F}\to\mathscr{G}$ (e.g. the homomorphism $\underline{\Bbb R}\to\Omega^0_M$ !).
It's also worth noting that by general nonsense (see Cartan-Eilenberg) the fact $\alpha$'s chain homotopy class is well-defined means we have a well-defined double-complex homotopy class of map $T_{\mathfrak{U}}\to T_{\mathfrak{V}}$; you don't need to worry about choices.
For $\phi$, it is also straightforward to see. $\Omega^\bullet_M\to T_{\mathfrak{U}}\to T_{\mathfrak{V}}$ is the map - at level $k$ - of $k$-forms $\omega$ to $(\omega|_U)_{U\in\mathfrak{U}}$ to $((\omega|_{\alpha(V)})|_V)_{V\in\mathfrak{V}}=(\omega|_V)_{V\in\mathfrak{V}}$, which is what we want. Abstractly, you can see this as the fact $\mathscr{F}\to\mathfrak{C}^\bullet(-,M;\mathscr{F})$ is a cone, viewing the right hand side as a functor of covers and refinements, for any sheaf $\mathscr{F}$ on $M$.
So we have commutative diagrams: $$\require{AMScd}\begin{CD}\check{H^\bullet}(\mathfrak{U},M;\Bbb R)@>\cong>>H^\bullet(\mathrm{Tot}(T_{\mathfrak{U}}))@<\cong<<H^\bullet_{dR}(M)\\@VVV@VVV@VV\mathrm{Id}V\\\check{H^\bullet}(\mathfrak{V},M;\Bbb R)@>>\cong>H^\bullet(\mathrm{Tot}(T_{\mathfrak{V}}))@<<\cong<H^\bullet_{dR}(M)\end{CD}$$
And we have compatibility in the composed horizontal isomorphisms. If, you know, you fix your isomorphisms to be exactly of this type and any sign-fiddling you do is done consistently.
