I have seen the proof of Prop 3.29 in Hatcher, Algebraic Topology which states for any non-compact connected manifold $M$ of dimension $n$, $H_n(M,R)=0$ for any ring $R$.
The proof essentially goes by realization of homology class realizing a global section of $M\to M_R$ where $M_R$ is formed by pair $(m,s_x)\in \cup_x M\times H_n(M,M-x,R)$ with $s_x$ comes from some $s\in H_n(M,M-B,R)$ where closed ball $B\ni x$ and this is topologized as sheaf space. Such a section must come from some n-chain of $M$ with compact image. In particular, such chain will have trivial homology in $H_n(M,M-y,R)$ with $y$ far away from image. And connectedness of $M$ dictates uniqueness of section for $M\to M_R$. It can be shown that any cycle is a boundary of a larger space containing the image. Hence every cycle is trivial in $H_n(M)$.
$\textbf{Q:}$ What is the intuitive reason to expect non-compact manifold connected $M$ $H_n(M,R)=0$ for any ring $R$?
$\textbf{Q':}$ I am not sure whether I have the following thoughts correct. It seems that every $n$ cycle is boundary in above setting.(And my guess is that there is no cycle but I am not sure about this.) Imagine there is an $n-$simplex in $M$. Clearly, it has non-trivial boundary. Now to remove boundary, I need to add more simplexes to cancel the boundary. However $M$ is non-compact, I need to add more and more simplexes. Since every chain is by definition of finitely generated by basis, there can not be too many simplexes inserted. Hence I have to conclude there is no cycles. Is this picture correct? I doubt about it as in Hatcher's proof, it went through showing this cycle must vanish as a homology class in a pre-compact set where I might doubt there might be non-trivial cycles in the pre-compact set.
The intuition comes from simplicial homology. If the connected manifold $M$ has the structure of a simplicial complex, and if $\sum_i a_i \sigma_i$ is a simplicial $n$-cycle in $M$, then it is easy to prove by induction that $|a_i|$ is independent of $i$. But if $M$ is not compact and if some $a_i$ is nonzero then one obtains a contradiction: the number of simplices is infinite, and all of the $a_i$'s are nonzero, but a simplicial chain must have only finitely many nonzero summands.