I think Anton said basically the same thing, but I'll expand a bit. When I think of an immersed submanifolds, two reasonable definitions come to my mind:
A map $f:N \to M$ such that N, M are both differential manifolds, $\dim M >\dim N$, and the map is locally an embedding, i.e. the derivative matrix at each point has no kernel.
The same as above, but with the additional requirement that the map be transverse to itself.
(In fact, for me an immersion is almost always number 2, but 1 might make more sense sometimes. In general, all books I've seen say that there is no universally agreed upon definition of immersed/embedded submanifolds)
I do not think it makes sense to think of the submanifold as just the image of that map. In particular, the main reason to have submanifolds is to talk about tangent vectors to the submanifolds, and this makes no sense unless you have the map. (When you imagine a tangent vector to the image, what you are actually taling about is a tangent vecotr to N).
If you accept 2 as the definition, it's an interesting question of whether you can reconstruct the map f from just the image in any reasonable unique way. I think the answer should be yes for reasonable examples, but there might be a weird counterexample. If 1 is the definition, the answer is certainly "no" (just imagine a figure eight where the self-intersection is a small interval rather than just a point). In any case, I don't think you'll be able to do anything with your immersed submanifold unless you have the map.
My answers to the specific questions of the original poster:
1) Union of countably many submanifolds is an immersed submanifolds iff you consider a disjoint union of countably many abstract manifolds a manifold. Note that for embedded submanifolds, it's always possible to construct a map from the corresponding abstract manifold to M.
2) This depends on whether you require the embedded submanifolds to be closed. A figure eight cannot be decomposed into a union of embedded closed differentiable manifolds. If they don't have to be closed, as Andrey said in the comments, you can cover N by open sets small enough that the map is an embedding on each.