 Solution

The two singly marked sides of the first triangle are glued with their endpoints at $A$ matched. A cross-section parallel to the remaining side has its two endpoints identified, becoming a circle; the apex becomes its cone point. Thus this quotient triangle is a disc, whose boundary is the doubly marked edge.

In the second triangle, the two triply marked sides run consecutively along the boundary from $E$ through $D$ to $F$. Pairing these sides in the indicated directions introduces one reversal; its boundary word is $aa{b}^{-1}$, where $b$ is the remaining doubly marked edge. This is a Möbius band with boundary $b$. One can obtain the usual rectangle model by cutting from the identified triple-edge seam to the midpoint of $b$ and opening the triangle.

The double markings glue the disc boundary to the boundary of that Möbius band. Capping a Möbius band by a disc givesThis description also verifies that the quotient is a closed surface, including the vertices. As a check, the vertex classes are $\{A\}$ and $\{B,C,D,E,F\}$; there are three edge classes and two faces. The Euler characteristic is $2-3+2=1$, consistent with the projective plane. The Möbius band exhibits nonorientability directly.