According to Topology by C Adams :
and according to Exercise 9.4. :
I understand the Example but I don't understand the case of $\mathbb R^2 - {\{O}\}$:
1- The only way to 'deform' $f$ to $g$ if there is a hole in origin is jump to the third coordinate. So $f$ and $g$ are homotopic in $\mathbb R^3 - {\{O}\}$ not in $\mathbb R^2 - {\{O}\}$. Am I right?
2- I don't understand the Hint, esp what is constant path and what is the relevance of path connectedness of $\mathbb R^2 - {\{O}\}$?
3- If $f$ and $g$ are still homotopic in $\mathbb R^2 - {\{O}\}$ so what is the explicit homotopy $F(x,t)$ in $\mathbb R^2 - {\{O}\}$ such that $F(x,0)=f(x)$ and $F(x,1)=g(x)$?
I am just a toddler in topology; detailed simple explanation would be much appreciated.
Your $(1)$ is incorrect.
For $(2)$, a constant path is a path that looks like $f(x) = (1,0)$ as a map from $I$ to $\mathbb{R}^2$. Path connectness gives directly that two constant paths are homotopic. In particular, you translate the entire path along the guaranteed path connecting the two images.
For $(3)$, you can write down the homotopy as a combination of two homotopies: retracting $f$ to the starting point and then the inverse of retracting $g$ to the starting point. An animated homotopy would look like $f$ shrinking back to the start, and then regrowing along the path of $g$.
More generally, lines are homotopic to their endpoints. This is often called being "contractible." Thus in path connected spaces, any two (simple, non-loopy) lines are homotopic.