Is the real projective plane flat? I am not sure if I am using the correct name, what I mean by the real projective plane is a flat square of space in which the upper edge is identified with the left/right reversed bottom edge and vice versa for the left and right edges. I would normally assume that it is flat, for the same reasons a torus is flat (I am based on intuition, I still  have zero  understanding of topology or differential geometry).
But if I look at what happens to an arrow moving up diagonally, like in the figure below, I find that angles are not preserved. I attempted to draw what happens with an arrow that moved diagonally and disappears on the upper right corner to reappear at the bottom left one. Or did I do something wrong?

 A: $\def\RR{\mathbb{R}}\def\PP{\mathbb{P}}$What people usually mean by flat is "having a metric with Gaussian curvature zero". Informally, this means that if you measure distances on your space, it locally looks like $\RR^2$. The OP sounds like someone who isn't familiar with these ideas, so I'm going to try to give a leisurely introduction. These issues were already raised in comments by Deane and Andrew D. Hwang.
Let $M$ be a surface with a metric and let $p$ be a point of $M$. Let $C_r$ be the set of points at distance $r$ from $M$, for $r$ small enough, this is a circle. Let $|C_r|$ be the perimeter of that circle. Then the curvature of $M$ at $p$ can be defined as
$$K(p) := \lim_{r \to 0^+} \frac{2 \pi r - |C_r|}{\pi r^3/3}. \qquad (\ast)$$
For a first example, if $M = \RR^2$, then $|C_r|  = 2 \pi r$ and the curvature is $0$. For a second example, if $M$ is a sphere of radius $R$ and $p$ is the north pole, then $C_r$ is the circle of latitude at angle $\tfrac{r}{R}$ from the pole, and $|C_r| = 2 \pi R \sin \tfrac{r}{R} = 2 \pi \left( r - \tfrac{r^3}{6 R^2} + \cdots \right)$ by the Taylor series of sine, so $M$ has curvature $\tfrac{1}{R^2}$.
By the Gauss-Bonnet theorem, for a compact surface $M$, we have
$$\int_M K dA = 2 \pi \chi(M)$$
where $dA$ is with respect to the area coming from the metric and $\chi(M)$ is the Euler characteristic. For example, the Euler characteristic of a sphere is $2$. The sphere of radius $R$ has area $4 \pi R^2$ and curvature $\tfrac{1}{R^2}$ as computed above, and we have $4 \pi R^2 \cdot \tfrac{1}{R^2} = (2 \pi) \cdot 2$, as predicted by the Gauss-Bonnet theorem.
In particular, if $K$ is identically zero, then $\chi(M)$ must be $0$. This happens for the torus and the Klein bottle, but not for $\RR \PP^2$; the Euler characteristic of $\RR \PP^2$ is $1$.

Okay, but what does this have to do with drawing squares with identified sides? Let's start with an easier question: Cut out the template below and glue the sides together. You'll get the surface of a cube, which is toplogically a sphere, and it will look flat. Why doesn't this violate Gauss-Bonnet?

If you use definition $(\ast)$ above, you will find that the curvature of your cube surface is $0$ at the interior of any face. You will even find that it is $0$ in the interior of every edge: If $p$ is in the interior of an edge, and $r$ is shorter than the distance from $p$ to either end of the edge, then you will find that $|C_r| = 2 \pi r$.
However, if $p$ is one of the eight vertices of the cube, then life is very different! In this case, $C_r$ is the union of three quarter arcs, so $|C_r| = \tfrac{3}{2} \pi r$. So the limit in $(\ast)$ blows up very quickly! Your cube surface has infinite curvature at the eight vertices, and $0$ everywhere else.
So, we need a version of Gauss-Bonnet for spaces with corners. For simplicity, let's do spaces like our cube surface, which are curvature zero everywhere except at finitely many points $v_1$, $v_2$, ..., $v_n$. Let
$$k_i = \lim_{r \to 0^+} \frac{2 \pi r - |C_r|}{r}$$
where $C_r$ is the circle of radius $r$ around $k_i$. Then
$$\sum_{i=1}^n k_i = 2 \pi \chi(M).$$
In our cube example, we get $8$ points where $k_1=k_2=\cdots=k_8 = \tfrac{\pi}{2}$. And $\tfrac{\pi}{2} \cdot 8 = (2 \pi) \cdot \chi(S^2)$, as predicted. (Exercise: Prove this formula for polyhedra! You will learn what Euler characteristic has to do with Euler's formula! This case is known as Descartes' Angle Deficit theorem.)
And this is what happens when you glue a square to itself as in your figure. The resulting surface has two corner points, coming from the $4$ corners of the square. The circle of radius $r$ around such a corner has perimeter $\pi r$; on the square, it looks like two quarter arcs in two of the opposite corners. So we have $\pi + \pi = (2 \pi) \cdot \chi(\RR\PP^2)$, as predicted.
A: 
Is the real projective plane flat?

No.  If is was flat it was a 2-torus or a Klein bottle.
More specifically, $P\Bbb R^2$ can't be endowed with a geometry that's flat everywhere (and for that matter, not even with a geometry that has constant curvature everywhere).  The 2-dimensional manifolds without boundary of constant curvature (in that sense) are:

*

*The 2-sphere (compact, orientable, positive curvature (i.e. it can be endowed with a natural geometry that has constant positive curvature everywhere))


*The Euclidean plane (infinite, orientable, flat (curvature = 0, i.e. it can be ebdowed with etc. etc.))


*Hyperbolic 2-space (infinite, orientable, hyperbolic (constant negative curvature))


*2-Torus (compact, orientable, flat)


*(Surface of) Bretzel-shapes with at least 2 holes (compact, orientable, hyperbolic)


*klein bottle (compact, non-orientable, flat)
