I am comfortable with de Rham cohomology, sheaves, sheaf cohomology and Cech cohomology.

I am looking to prove the following theorem:

If $M$ is a smooth manifold of dimension $m$, then we have the following isomorphism for each $k \leq m$ $$ H^k_{\text{dR}}(M) \cong \check{H}^k(M; \mathbb R_M). $$ By $\mathbb R_M$ I meant the constant sheaf of $\mathbb R$ on $M$.

I'll outline the proof I am following and give my two questions as they come up.

If we let $\Omega^k$ denote the sheaf of germs of $k$ forms on $M$ then by the Poincaré lemma we have an exact sequence $$ 0 \rightarrow \mathbb R_M \rightarrow \Omega^0 \rightarrow \Omega^1 \rightarrow \ldots \rightarrow \Omega^m \rightarrow 0, $$ with differential maps being the exterior derivative.

Why is this sequence exact but the de Rham complex not?

In particular, I think I may not be clear on how a "sheaf of germs" is different from a sheaf? And if I took global sections would I get back the usual de Rham complex?

Anyway, assuming that the sequence is exact, we therefore get a series of short exact sequences (SES) of the form $$ 0 \rightarrow d\Omega^{k-1} \rightarrow \Omega^k \rightarrow d\Omega^k \rightarrow 0. $$ The sheaf $\Omega^k$ is fine, and hence its cohomology $H^i(\Omega^k)$ vanishes for $i>0$. Hence we get an isomorphism $H^i(d\Omega^{k-1}) \cong H^{i+1}(d\Omega^k)$ for each $i$. However, the last sentence of this proof is a mystery to me.

"At one end of the chain is the Čech cohomology and at the other lies the de Rham cohomology."

I can kind of see why after taking global sections we get back de Rham cohomology - i.e. the $H^0$ that come from the SES above is $H^k_{\text {dR}}(M)$. And the Cech cohomology coincides with the sheaf cohomology. But I am not sure how to relate them: should I be varying $k$ and get isomorphisms across the long exact sequences corresponding to different $k$?