I have a $1 \times 1$ square and I cut the triangle that has vertices at the center and bottom right and left corners of the square. Then I glue this triangle to the right edge of the square as shown in the picture:
The task is to cut the second shape in two such that after rotating and maybe flipping those two shapes, we get the unit square back. The cut can be any curve that separates the shape into two pieces. For example, the cut shown by the color green trivially gives a square after we rotate the triangle back into the place where we cut it from. The question is whether there are other cuts, and if they exist whether they are finite or infinite. 
By arguing that a distance larger than $\sqrt2$ cannot exist, I managed to reduce the possible curves to the ones starting at the Yellow line and ending at the Red zigzag as shown in the image above. But I couldn't go further. Also, a continuation of this problem would be to show the existence or finiteness of solutions given any regular polygon.
We can show that any working straight or zigzag (alternating concave-convex angle) cut divides the piece into two pieces A and B, where A has an convex angle a and B has an convex angle b such that $a+b=180^{\circ}$.
Let's start by forgetting about the square specifically. Instead, we wonder whether this shape can be divided into two pieces that can be formed to make any quadrilateral in any way but the trivial case. We notice that for almost all convex $a$-gon A and convex $b$-gon B where A has no angle $a$ and B has no angle $b$ such that $a+b=180^{\circ}$, when glued together we get a polygon with at least $(a+b)-2$ vertices. For example, here's two 5-gons joined together into an 8-gon.
However, we're assuming that when joining two convex vertices, we always form another vertex. This, however, is not always true, as if we join two pieces A and B, where A has an convex angle a and B has an convex angle b such that $a+b=180^{\circ}$, the angles a and b may combine to "cancel out", so to speak, meaning that both vertices "disappear" from our count. Let us assume that this condition does not happen for the rest of our analysis.
We need to place a similar lower bound on the number of vertices in the gluing of concave polygons. For the sake of terminology, let us define an $(a,b)$-gon as a simple polygon with $(a+b)$ vertices, $a$ of which are convex and $b$ of which are concave. It is evident that in an extremely optimal (and potentially impossible setting), when we join an $(a,b)$-gon and a $(c,d)$-gon, we might be able to get every concave angle in one polygon to "cancel out" a convex angle in the other. Thus, we would obtain a polygon with at least $(a-d)+(c-b)-2$ vertices, subtracting the $2$ again as before. You can of course rearrange this formula, but I think this representation is most intuitive. For example, to make a 4-gon from the blue $(5,2)$-gon below, you need to glue on the red $(6,3)$-gon.
This formula allows us to show that dividing our shape with zigzag rather than straight lines has no effect on the number of vertices in the polygons we can ultimately create. Adding zigzags in a straight line increases the number of convex vertices in one polygon, but simultaneously increases the number of concave vertices in the other polygon by the same amount. These cancel, so even in our hyperoptimal (and again, probably generally impossible) case, the minimum vertices of the glued polygon remains the same. Here's an example: no matter how many zigzags we add to this dividing line, the resulting blue and red pieces can't be joined together to make anything with less than 6 vertices.
Now, we can analyze our particular shape using casework, building off of OP's result that the line must start at the top yellow line (but not at the top left corner, because then we'd have lines longer than $\sqrt{2}$) and come down to the red line (but not the bottom left corner for the same reason). Because we don't care about distance, only the vertex counts of the resulting polygons, many points are interchangeable. Precisely, we only have seven classes of points - left half of yellow line, right half of yellow line, top right corner, left half of red line, midpoint of red line, right half of red line, bottom right corner. Note: Zigzag cuts alternate between concave and convex, causing this cancellation. If one drew a cut with, say, three concave angles (from the perspective of the left piece), these would not apply. It may be possible to reduce these further, but this is good enough for the casework. Remember that zigzags don't affect the final vertex count, so we can add them to our lines if needed.
We see in the cases above that the only line that produces polygons that may form a quadrilateral is in the trivial case, from the top right corner to the bottom right corner. Thus, assuming our initial condition regarding angle sums, no non-trivial solution exists. As OP has mentioned in the comments below, though, it is possible to find a non-trivial solution, but as shown, there must be two angles (one in each piece) that sum to 180 degrees.
Note: It may be possible to generalize from our piecewise-linear cuts to smooth curves by taking limits. However, this may break our method of counting, so I won't extend to that here.