Consider a manifold $ N $ defined as follows $$ N=\{n\otimes n-m\otimes m:n,m\in S^2,\,\,(n,m)=0\}\subset M^{3\times 3}, $$ where $ S^2 $ denotes the two dimensional sphere, $ (\cdot,\cdot) $ represents the inner product of the space $ \mathbb{R}^3 $ and $ M^{3\times 3} $ denotes the set of $ 3\times 3 $ matrices. For a cloase curve $ \alpha:S^1\to N $, we can consider the homotopy class of it. Define maps $$ \pi_1:N\to\mathbb{R}P^2,\,\,\pi_2:N\to\mathbb{R}P^2 $$ as follows. For any $ A\in N $, we can represent $ A=n\otimes n-m\otimes m $ with $ (n,m)\in S^2\times S^2 $ and $ (n,m)=0 $, we define $ \pi_1(A)=n\otimes n $ and $ \pi_2(A)=m\otimes m $. I guess that if the homotopy class of $ \alpha $, denoted by $ [\alpha] $, is non-trivial, then there is one of $ [\pi_1(\alpha)] $ and $ [\pi_2(\alpha)] $ non-trivial. However I don not know how to prove this or give a conterexample. I have already know the covering space and fundamental group of $ N $ but I do not know how to go on. Can you give me some hints or references?
A further question is that if $ [\alpha] $ corresponds to the element of order $ 4 $ in the fundamental group, is such result true?
This is not true.
There's an action of $SO(3)$ on such pairs $(n, m)$ which for $O\in SO(3)$ sends the pair $(n, m)$ to $(On, Om)$. The action is transitive. You've correctly identified that the stabilizer of $(n, m)$ seen as a point on $N$ is a subgroup isomorphic to $\Bbb{Z}/2\Bbb{Z}\times\Bbb{Z}/2\Bbb{Z}$ with one generator sending $n$ to $-n$ and the other generator sending $m$ to $-m$. The manifold $N$ is a quotient of $SO(3)$ by a $2\times 2$ discrete subgroup. However, $SO(3)$ is homeomorphic to $\Bbb{R}P^3$ and is not simply connected. It has fundamental group $\Bbb{Z}/2\Bbb{Z}$.
For a choice of $(n, m)$ let $k=n\times m$ (the cross product). $k$ completes $(n, m)$ to a right-handed orthonormal basis. Now let $R_\theta(k)$ be the rotation in $SO(3)$ around axis $k$ by angle $\theta$. As you trace a loop in $SO(3)$ letting $\theta$ range over $2\pi$, in the manifold $N$, $(n, m)$ goes to $(-n,-m)$ at $\theta=\pi$ which is the same as $(n, m)$ in $N$, then after another turn of $\pi$ you get back to $(n, m)$. At this point the images of the loop in $\Bbb{R}P^2$ in your maps $\pi_1, \pi_2$ is a trivial loop, because you've gone around twice. In $SO(3)$ you've done one turn of $2\pi$, but that is not a trivial loop in $SO(3)$ because $SO(3)$ is not simply connected. This loop is the generator of the fundamental group of $SO(3)$.
The correct thing is that $N$ is homemorphic to $SU(2)$ divided by the quaternion group $\{\pm 1, \pm i, \pm j, \pm k\}$. $SU(2)$ is simply connected and is homeomorphic to $S^3$. Then after dividing by the center of the quaternions which is $\{\pm 1\}$ you get $SO(3)$. The quaternions divided by its center leaves you with a group which isomorphic to $2\times 2$, and then dividing by that you finally get $N$.