I had to think about this a while to untangle the issue.
First, Bott/Tu have already restricted to a compact disk bundle, so the homotopy is proper. But, second, I believe we don't even need that. If you look at the construction of the chain homotopy (around p. 34), the form $\tau$ is given in our case by $\tau = KH^*\omega$, where $K$ is integration over the fiber of the homotopy $H$. Since $\omega$ has compact support in the base manifold, so will $\tau$.
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
Your question is not very precise, but I would say the answer is no. Consider $M$ to be the $0$-manifold given by the singleton. Take the simplicial complex given by the simplest triangulation of $[0,1]$ (two vertices, one edge), which satisfies $M \simeq [0,1]$. Then the cellular complex of $K$ is $C_0(K) = \mathbb{F}_2^2$ and $C_1(K) = \mathbb{F}_2$. There is no triangulation $K^\vee$ such that $C^0(K^\vee) = \mathbb{F}_2^2$ and $C^{-1}(K^\vee) = \mathbb{F}_2$.