= Solution
Treat the two squares as a single edge-pairing construction, including the identifications between different squares. Number the corners of the left square counterclockwise from its lower left corner by $A,B,C,D$, and similarly use $E,F,G,H$ on the right. The single arrows identify $D$ with $H$ and $C$ with $E$. The double arrows identify $G$ with $F$ and $H$ with $G$. The triple arrows identify $A$ with $E$ and $B$ with $F$. The quadruple arrows identify $A$ with $C$ and $D$ with $B$.
Thus there are precisely two <vertex> classes, $\{A,C,E\}$ and $\{B,D,F,G,H\}$. There are four <edge> classes and two two-dimensional cells, so the resulting closed connected <surface> has <Euler characteristic>
$$
\chi=2-4+2=0.
$$
It is a <nonorientable surface>: on the left square the quadruple-arrow sides have the same orientation relative to a traversal of its boundary, giving a twisted band after gluing. Equivalently, no choice of orientations for the two faces makes the induced directions opposite on every paired <edge>. The <vertex> links are circles, so these <edge> identifications produce a closed <surface>, rather than a singular cell space. The <classification of compact surfaces> now gives
$$
\boxed{\text{Klein bottle}.}
$$
A <torus> also has <Euler characteristic> zero, but is excluded by the failed orientation test.
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