This is not a conclusive answer, but here's how I would approach this.
Let $\Delta_n(M)$ be the Abelian group of singular $n$-chains and $\Delta^n(M;\mathbb{R})=\mathrm{Hom}_\mathbb{Z}(\Delta_n(M),\mathbb{R})$ the $\mathbb R$-valued signular $n$-cochains. There are two subcomplexes that are relevant to the question: the complex $\Delta_*^\infty(M)$ of smooth singular chains (as explained in Bredon's book, for example), and the complex $\Delta_c^*(M)$ of compactly supported cochains, i.e. singular cochains that vanish on all chains with image outside of a compact set (which depends on the cochain).
Now, without having a reference or a proof, I would bet some money that the inclusion $\Delta_*^\infty(M)\hookrightarrow\Delta_*(M)$ is a chain homotopy equivalence. This should, in turn, dualize to chain homotopy equivalences
$$\Delta^*(M;\mathbb R)\to\Delta^*_\infty(M;\mathbb R) \quad\text{and}\quad \Delta^*_c(M;\mathbb R)\to \Delta^*_{\infty,c}(M)$$
where $\Delta^n_\infty = \mathrm{Hom}(\Delta_n^\infty,\mathbb R)$ and $\Delta^n_{\infty,c}$ is the compactly supported analogue.
Next, integration gives rise to chain maps
$$\Psi\colon \Omega^*(M)\to \Delta_\infty^*(M) \quad\text{and}\quad \Psi_c\colon \Omega^*_c(M)\to \Delta_{\infty,c}^*(M).$$
Bredon proves that $\Psi$ induces an isomorphism on cohomology and it should be possible to adapt the proof to show the same for $\Psi_c$.
Finally, the cohomology of $\Delta_c^*(M)$ is known a singular cohomology with compact supports, denoted by $H^*_c(M;\mathbb R)$. If $M$ has an orientation, then Poincaré duality gives isomorphisms
$$H^{n-i}_c(M;\mathbb R)\cong H_{i}(M;\mathbb R)$$
with singular homology on the right hand side. And if everything above goes through as claimed, then the left hand side is isomorphic to compactly supported de Rham cohomology $H^{n-i}_{dR,c}(M)$.
I'll try to find some references later. Other duties are calling.
Best Answer
If you think about singular homology, then the support is quite easy to understand : for a chain $\sum n_i\sigma_i$ where the $n_i$ are non zero integers, the support is simply the union of the images of $\sigma_i$. Now since the standard simplexes are compact, so are their images and so is a finite union. It follows that the support of a cycle in simplicial homology is compact. Note however that the support of two homologous cycles are not necessarily identical.
The same hold Borel-Moore cycles. This is the union of of the images of all simplexes. The fact that it is locally finite implies that the support is closed, but not necessarily compact.
About terminology, Borel-Moore homology is sometimes called homology with compact support. This historical terminology is wrong, this is really the usual homology which has compact support. This is clear with the notion of support above but also in the light of Poincaré duality. Indeed, you have isomorphisms $H_i(X)\simeq H^{d-i}_c(X)$ and $H_i^{BM}(X)\simeq H^{d-i}(X)$ which also shows that homology has compact support (since it is isomorphic to cohomology with compact support) and Borel-Moore homology has non compact support (since it is isomorphic to cohomology). You also have intersection pairings $$H_i\otimes H^j\to H_{i-j}$$ $$H_i^{BM}\otimes H^j\to H_{i-j}^{BM}$$ $$H_i^{BM}\otimes H^j_c\to H_{i-j}$$ $$H_i\otimes H^j_c\to H_{i-j}$$ You can see that the intersection of something (cycle or cocycle) compact with something non-compact gives a compact cycle as expected (see the third for example). Only two give rise to perfect pairing : the first and the third, that is when we pair something compact with something non-compact. This is well known in functional analysis.