Tarjan's SCC : example showing necessity of lowlink definition and calculation rule?
For a given vertex, the only thing that matters in the algorithm is if there is an edge from the vertex or a child-vertex to a ancestor-vertex in the DFS exploration tree. In this case, we know that the vertex belongs to the same SCC as it's parent. In such a case, we are sure that the vertex itself or a child-vertex has a direct back edge to an ancestor of the vertex, and this ancestor has a lower index that the current vertex, so the v.lowlink will differ from v.index. We actually do not need to know the top-vertex of the SCC at this step.

So, using v.lowlink  := min(v.lowlink, w.lowlink) instead of v.lowlink  := min(v.lowlink, w.index) will lead to a correct algorithm too.
The difference is that in this case, v.lowlink will not represent the vertex with the smallest id reachable from v  nor the SCC top vertex, but just v if there is no such vertices, or any parent vertex that belongs to the same SCC (maybe the top-vertex), depending of DFS order. I think it's just lack in formal definition, and thus, wasn't used.

Let's see what appends when we use this alternative instruction. When we update v.lowlink because we have found a new back edge to a ancestor vertex, we will now set v.lowlink with the value of w.lowlink. But this vertex is an ancestor and is still in the stack: we may not have already explored all the child vertices of this ancestor and thus, w.lowlink can change in the future: this is why we dont's know exactly what will be the final value of v.lowlink because it depends of the DFS order. We can juste be sur that it will be w.index or a vertex with a lower index that belongs to the same SCC (so the algorithm will work as expected): the value of lowlink still differs for differents vertices of the same SCC.

The following example show how DFS order alters lowlink values in this alternative algorithm:

In this case, assuming we are exploring vertices in the order 0-1-2-3-4, when we consider the vertex 3, the vertex 4 has not been explored yet. The lowlink value of vertex 1 is so 1, and vertices 2 and 3 will take the same value. But we will later find the back edge from 4 to 0, and set the lowlink value of vertices 4 and 1 at 0. If we had explored vertex 4 prior to vertex 2, all vertices would have 0 as lowlink value.

And for the last question, yes, all this verteces belong to the same SCC, rooted by x.lowlink or an ancestor of the vertex with id x.lowlink