The question I am trying to answer is: Does the quotient map $ q:[0,1] \times [0,1] \to \mathbb R \mathbb P^2$ extend to a covering map $\mathbb R^2 \to \mathbb R \mathbb P^2$
I know that the quotient map $ [0,1] \times [0,1] \to T$ extends to a covering map $\mathbb R^2 \to T $ but I'm not even sure how to approach the new question (as I thought it would be similar).
For a start I don't know how to extend $q$ onto $\mathbb R^2$ because the way I'm thinking about it is by identifying opposite edges of a square, whereas for the Torus I'm using exp to define my covering map.
I also know that the quotient map $p: S^2 \to \mathbb R \mathbb P ^2$ is a covering map but not sure how this would help when looking at the quotient map above.
Any help/hints would be great, thank you.
There is no cover from $\mathbb{R}^2$ to $\mathbb{R}P^2$ because the universal cover of $\mathbb{R}P^2$ is $S^2$ which is unique up to homeomorphism.
For the particular case you are thinking about, extending the quotient map from the square to a covering map from $\mathbb{R}^2$ to $\mathbb{R}P^2$ by "fitting" a infinitely many squares into the plane, note that this is would be an infinite sheeted cover and hence not possible for algebraic reasons.
If such an infinite sheeted cover $p:\mathbb{R}^2 \to \mathbb{R}P^2$ existed then $p_*(\pi_1(\mathbb{R}^2)) = \{0\} \subset \pi_1(\mathbb{R} P^2) = \mathbb{Z}_2$ would be of infinite index. But of course $\mathbb{Z}_2$ has order $2$ so this is impossible.