= Solution
The surface is a \b[<Klein bottle>]. Here is a cell-count justification that keeps all the arrow directions. Name the corners of the first square $a,b,c,d$, beginning at its upper left and proceeding clockwise; name those of the second $e,f,g,h$ in the same way. The single-arrow gluing identifies $a$ with $e$ and $b$ with $h$. The triple-arrow gluing identifies $d$ with $h$ and $c$ with $g$. The first square's quadruple-arrow gluing identifies $b$ with $d$ and $c$ with $a$. The second square's double-arrow gluing identifies $e,f,g$ with one another. There are exactly two resulting vertex classes:
$$
\{a,c,e,f,g\},\qquad\{b,d,h\}.
$$
There are four edge classes and two faces, so the <Euler characteristic> is $2-4+2=0$. Every edge is paired and the corner links close into circles, giving a <compact> surface without boundary; the cross-square identifications make it <connected>. It is nonorientable: the quadruple-arrow identification glues two boundary edges of the same square in the same boundary-traversal direction, so no <orientation> of that face can extend across the gluing. The classification of <compact> <connected> surfaces now identifies the nonorientable surface of <Euler characteristic> zero as the <Klein bottle>.
Geometrically, the internal identification in each square produces a <Möbius band>. The remaining single- and triple-arrow edges comprise their two boundary circles, which are glued together. The double of a <Möbius band> is another description of the same <Klein bottle>.
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