M is an ndimensional closed orientable manifold. I find in a book "Intuitively,the fundamental class can be thought of as the sum of the (topdimension) simplices of a suitable triangulation of the manifold". My explanation is we can use the Mayervietoris sequence to "glue" the local orientations inductively over a finite oriented atlas together to construct the fundamental class. One open subset of the atlas correspond to one simplex. Am I right?

Here is how I would prove it. You need to know two things. 1. If $M$ is connected, then $H_n (M) \to H_n (M,Mx)$ is an isomorphism (proven in any appropriate textbook). 2. If $\sigma:\Delta^n \to R^n$ is an embedding that has $o$ as an interior point of its image, then it represents a generator of $H_n (R^n,R^n  o)$. This is easily reduced to showing that the identity map on $\Delta^n$ represents a generator of $H_n (\Delta^n, \Delta^n  o)=H_n (\Delta^n, \partial \Delta^n)$, where $o$ is an interior point. By long exact sequence and excision, $H_n (\Delta^n, \partial \Delta^n) \to H_{n1} (\partial \Delta^n) = H_{n1}(\partial \Delta^n, \Lambda^n) \cong H_{n1} (\Delta^{n1}, \partial \Delta^{n1})$, where $\Lambda^n$ is the horn, i.e. the union of all but one boundary simplices. By induction, the claim follows. To get back to the manifold: pick a point $x$ in the interior of $\sigma$, one of the topdimensional simplices in your triangulation. All that remains to be proven is that the sum of all topdimensional simplices maps to $\sigma$ under $H_n (M) \to H_n (M,Mx)$, but that follows directly from what I wrote. 


Assuming you mean a smooth manifold, taking the top exterior power of the tangent bundle at every point gives you the determinant bundle, which is trivial when the manifold is orientable. Therefore you get a global nowhere vanishing section, which is an orientation. Along the same lines, one can produce a nowhere vanishing topdimensional "volume" form which of course defines a fundamental class in de Rham cohomology. These are all correct "intuitions" but in your context you don't seem to be working with de Rham cohomology but rather with some version of singular (or simplicial) homology. It seems to me that your intuitions are all correct though of course without exact definitions it is hard to be sure. Certainly keeping de Rham's theorem in mind clarifies the "global" picture. 

