Question: How to quickly and clearly argue/show that there **is** or **is not** a linear surjective map $\phi$

$\phi: \mathbb{R} \to \mathbb{R}²$

Answer: 
Assuming you mean an $\mathbb{R}$-linear map, one can argue as follows: let $v = \varphi(1)$. Then for all $\alpha \in \mathbb{R}$, $\varphi(\alpha) = \varphi(\alpha \cdot 1) = \alpha \varphi(1) = \alpha v$. Thus the image of $\varphi$ consists of all scalar multiples of the vector $v$. This is the origin if $v = 0$ and a line through the origin otherwise. It is clearly not the entire plane: e.g., if $v = (x,y)$ is not zero, then $(-y,x)$ is not of the form $\alpha (x,y)$ for any $\alpha \in \mathbb{R}$.
For a linear transformation in this case we have:
 1=dim R =dim ImT + dim KerT.
Then dim ImT is 0 or 1.