Why can't we approximate the General TSP while we can approximate the Euclidean TSP? Euclidean TSP is approximatable, whereby the triangle inequality is obeyed. However, what is the exact reason which does not allow us to approximate General TSP?
 A: I assume you mean Travelling Salesman Problem. It's not clear what you mean by 'the exact reason', though.
The way we know approximating the general travelling salesman problem to within any constant factor is NP-hard, is that access to an approximation algorithm would allow one to solve the Hamiltonian Cycle Problem, which is NP-complete. This implies that no polynomial-time approximation algorithm is known (since that would imply P=NP), but also that there is no proof that one could not exist (since such a proof would imply P!=NP)
The reason you can't generalise the Euclidean TSP 2-approximation algorithm further than metric TSP is that the proof relies strongly on the triangle inequality.  The proof starts with finding the minimum spanning tree (which is easy). We then note that a path tracing out the minimum spanning tree and returning to the starting point is a tour, and that it has length twice the length of the minimum spanning tree.  So far, you can do this for the general TSP and get a usable tour.
For points in Euclidean space or a metric space, the minimum spanning tree has total length no greater than any tour, by the triangle inequality. This means the tour based on traversing the minimum spanning tree twice has at worst twice the length of the optimal tour.
For non-Euclidean graphs there could (in principle) be some alternative tour that's much shorter than the one you get from the minimum spanning tree, so there's no (known) bound on how bad the minimum spanning tree approximation algorithm can be.
