No, the idea is that they are all vectors in the same all-encompassing vector space. You can always sum them. What's missing on the formula is that all of these vector spaces are subspaces of the same larger one. That's implicit, otherwise the sum wouldn't make sense. Think that you are playing inside R^n and juggling a few lower-dimensional subspaces of it.

You can think of each vector space as being embedded in the direct sum of the two.

That makes sense, thanks!