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.
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.
Best Answer
Yes they are the same and what you call the trace map (and its dual in cohomology) is often called the transfer homomorphism. I will generalize to $X$ and $Y$ oriented triangulable closed $n$-manifolds but assume that $f : X \to Y$ is a degree $d$ covering map to ensure that the basepoint does have $d$ preimages. With this assumption it is easy to check that the transfer map on chains commutes with taking boundaries, so does produce a map on homology. With some more care you can also make it work for branched covers. Denote the transfer or trace map by $\DeclareMathOperator{\tr}{tr} \tr : H_*(Y) \to H_*(X)$.
Let $\alpha \in H^k(Y)$. Note that if $\sigma \in C_k(Y)$ and $\sigma'$ is any lift in $C_k(X)$ then by definition $f_*(\sigma') = \sigma$, so $f^*(\alpha)(\sigma') = \alpha(\sigma)$. From this it follows that for $h \in H_*(Y)$ (possibly assuming $* \ge k$ depending on how general your definitions are), $$\tr(h \frown \alpha) = \tr(h) \frown f^*(\alpha).$$ Taking $h = [Y]$ (the fundamental class of $Y$) you get $$\tr[Y] \frown f^*(\alpha) = \tr([Y] \frown \alpha).$$ This is exactly what you want once we verify that $\tr[Y] = [X]$. But if you start with a triangulation of $Y$ and pull back to $X$ then both $\tr[Y]$ and $[X]$ are just the sum of all the oriented top-dimensional simplices in $X$.
As a remark, one advantage of the Poincaré duality definition of this wrong-way or Umkehr map is that it works for any continuous map of oriented closed manifolds, or even more generally any proper continuous map of oriented manifolds.