I was told that a quotient space of a closed annulus centered at the origin obtained with a relation $x \sim - x$ for $x$ in the boundary is homeomorphic to a Klein bottle, which is a connected sum of two real projective planes.
So if my quotient map is $q$ and $U$ is the annulus without the inner boundary and $V$ is the annulus without the outer boundary, I was told $q(U)$ and $q(V)$ are projective planes missing a point, where the missing point is $q(\text{missing boundary})$.
However, I do not see this directly, and not sure how to show this.
Also, once I show this, I can see what is the fundamental group of the space by the consequence of the Van Kampen theorem, but what would be the generators of the fundamental group?
If you can appeal to the classification of connected, compact surfaces, then you can look at a cell decomposition: vertices are the 4 (really only 2) points where the boundaries meet the $x$-axis, edges are the 4 arcs (really only 2) plus the 2 segments along the $x$-axis, and 2-cells are the upper and lower regions these enclose. You can compute the Euler characteristic and can directly find an embedded copy of a Möbius band.
For the statements about the quotient maps, it might be easier to understand what's going on if you replace "are" by "are homeomorphic to". A surface minus a point is homeomorphic to a surface minus a closed disk centered at this point. This point of view reveals the boundary.
For the fundamental group, since you are thinking about the Van Kampen theorem, a natural choice would be to look at a generator for the fundamental group of the intersection of $U$ and $V$ and generators for the fundamental groups of $U$ and $V$. This will give three generators, but you can eliminate one using the relations that arise from Van Kampen's Theorem. A little bit of care is needed in choosing a base point.