Draw a circle on the surface and orient it. If the surface is non-orientable then there's a way to take a long walk, return to the circle, and find that the direction of the orientation has changed.
If the surface is non-orientable then I can draw a very small circle, put an orientation on it, then there's a path I can take that brings me back to find the orientation has changed.
Let's be more explicit.
Take a non-self-intersecting embedding of RP^2 in R^4, a point P in our RP^2, and a sufficiently small epsilon e. Then take three points on the circle of size e centred on P, and think of them as going in order, thus defining an orientation. Now take an appropriate walk around RP^2 and return to the circle. For an appropriate walk I will now find that the order of the points on my circle has reversed.
That's a more precise way of saying what I originally intended, and in that context your comment doesn't make sense to me. Can you expand on it?
ColinWright's point is that drawing such an arrow on a tiny circle is equivalent to drawing the letter R on the 2D manifold. It gives a local orientation to the 2D surface. (If you wish you may think of this as an arrow into some 3D embedding space, but you don't have to.)
If the 2D space is orientable, then when you take a copy of this little circle (or letter R) and go for a long walk, when you get home your copy will always match the original. That's all that orientable means. In the standard usage, it's a property of the 2D manifold, not of the particular walks you take. I think this is the point of confusion here.