All of my geometric intuition for "immersed" versus "embedded" surface is contained in my geometric intuition for "immersions" versus "embeddings". In particular, as many people have pointed out, immersions need not be injective. But, of course, even injective immersions need not be embeddings. As a very simple example, consider the map $f : (-\pi, \pi) \to \mathbb{R}^2$ given by
$$
f(t) = (\sin t, \sin 2t).
$$
The image of this map is a sort of "figure 8" in the plane, traced out starting at the origin, moving through quadrants II, III, I, and IV, in that order, as $t$ moves from $-\pi$ to $\pi$. It's easy to see that $f$ is an injective immersion, but $f$ is not an embedding, since every neighborhood of $\mathbb{R}^2$ containing the origin also contains points of the form $f(-\pi + \epsilon)$ and $f(\pi - \epsilon)$ for all sufficiently small $\epsilon$.
As another one-dimensional example of this type, you could consider the closed topologist's sine curve with a loop, which is the graph of $g(x) = \sin(1/x)$ for $x \in (0, 1]$ together with the $y$-axis between $y = -1$ and $y = 1$ together with a "loop" smoothly connecting the point $(0, -1)$ to the point $(1, \sin(1))$. It's clear that there is some injective immersion $f : [0, \infty) \to \mathbb{R}^2$ whose image is this curve, and this immersion is not an embedding.
You can, of course, easily make either of these example into a surface by considering $h: (-\pi, \pi) \times (0,1) \to \mathbb{R}^3$ given by $h(t, s) = (f(t), s)$.
On the other hand, I don't really know any examples of injective immersions of surfaces which aren't embeddings that are "interesting" in a way that's fundamentally different from the examples above. The idea I have is that immersions are allowed to "approach themselves" or "limit onto themselves" in crazy ways that embeddings are not. In particular, if $f : X \to Y$ is an injective immersion, the topology on $f(X)$ as a subspace of $Y$ might be very different than the topology on $X$.
- The notion of a local diffeomorphism makes sense only if the domain and the range are smooth manifolds. If the image of a map happens to be a smooth submanifold of the target manifold, one can say " $f$ is a local diffeomorphism onto its image" by restricting the range. Any other use is just made-up (by various MSE users, it seems) and should be avoided (at least until you are very comfortable with the subject). Instead you can simply say:
...The image of a map $f: X\to Y$ is a smooth submanifold and $f: X\to f(X)$ is a local diffeomorphism.
You may also sometimes encounter the following, describing what an immersion is:
A map $f: X\to Y$ of smooth manifolds is an immersion if and only if locally, it is a diffeomorphism to its image, meaning that $\forall x\in X \exists$ a neighborhood $U$ of $x$ such that $f(U)$ is a smooth submanifold of $Y$ and $f: U\to f(U)$ is a diffeomorphism.
But, again, given ambiguity of the language, it is better to avoid using this terminology in the beginning. The ambiguity comes from the word "image": It can either mean the image of the original map or the image of the map with the restricted domain.
- Everything that you wrote up to the line "However, the first and third posts..." is correct and proofs are very straightforward.
However: I did not check your guesses on how it may or may not be related to various MSE posts.
One thing, you should not repeat ad nauseum "with dimension". (Every manifold has dimension and, except for the empty set, its dimension as a smooth manifold equals its dimension as a topological space. As for the empty set: For every $n\ge 0$, the empty set is a manifold of dimension $n$. At the same time, from the general topology viewpoint, the empty set has dimension $-1$.)
- As for what various MSE users meant in their answers and comments, I prefer not to discuss: Frequently, there is no consistency in their use of mathematical terminology. (Many are only beginners, many have trouble with English, etc.)
Addendum. I am not sure who came up with the idea of allowing manifolds to have variable dimension on different connected components, but I wish this never happened as this just leads to a confusion. I checked several sources in geometry and topology, and the only author allowing manifolds to have variable dimension is Lang.
Best Answer
The map $\phi$ is not a homeomorphism onto its image $S\subset{\mathbb R}^3$. The surface $S$ is an infinite vertical cylinder that intersects the $(x,y)$-plane in a curve which is part of a folium of Descartes, see the following figure:
This curve does not have a self-intersection, but almost. The inverse map $\phi^{-1}$ is well defined, but is not continuous at the points $\phi(0,y)=(0,0,y)$. Any neighborhood of $(0,0,0)$ contains points of $S$ that are mapped by $\phi^{-1}$ onto points in $(-1,\infty)\times{\mathbb R}$ of the form $(M,0)$ with $M\gg1$, hence lying far away from $\phi^{-1}(0,0,0)=(0,0)$.