1 and 2: yes.
3: let us take first the case of a $(0,0)$ tensor, i.e. a function $f:M\to\mathbb R$. In this case we are given that $v(p)f=0$ for all $p\in M$ and are asked to show that for every $p\in M$, $g(t):=f(\phi_t(p))$ is constant. Now calculate that $g'(t)=v(\phi_t(p))f=0$, hence $g$ is constant.
(Note that the calculation uses the group property of $\phi_t$, i.e. $\phi_{t+s}=\phi_t\circ\phi_s$).
For a general tensor apply the above argument to each of its components wrt some basis.
It's a bit sketchy but I hope you can fill in the details.
- 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
A diffeomorphism is typically presented as a smooth, differentiable, invertible map between manifolds (or rather, between points on one manifold to points on another manifold). For example, take two sheets of paper and curl one of them up. There exists a diffeomorphism that relates points on the two sheets.
It sounds like you might be learning about Killing vectors. Changes in coordinates can be considered diffeomorphisms--instead of a passive relabeling of points, you are actively deforming spacetime into another shape, but one that is changed only by the coordinate transformation. Usually, tensors follow a strict transformation law under coordinate system transformations, but Killing vectors correspond to a symmetry in which the transformation law yields no change. A good example would be translational symmetry in Euclidean space. You can move a system any way you like, and aside from the points being relabeled, the fields themselves don't change.