Explain a mapping? Let $Z = \{(a, b) : a \ge 0 \text{ and } b = 0, \text{ or } a = 0 \text{ and } b \ge 0\}$.
Find a $C^\infty$ smooth function $\phi : \mathbb{R} \to \mathbb{R^2}$ which is $1$−$1$ and onto $Z$.
Q) Can someone please explain what $\phi$ is?
I think it's a tangent vector to my function evaluated with respect to or dependent on time?

The $\phi$ in question needs to be a function.  If we think about the input parameter as "time", then $\phi$ defines a trajectory in $\Bbb R^2$.  Our function needs to be 1-1, which means we can't pass through the same point twice, and it needs to be onto $Z$, which means the path needs to cover all of $Z$.
A function that almost works, but fails to be differentiable, is
$$
f(t) = \begin{cases}
(|t|,0) & t < 0\\
(0,t) & t \geq 0
\end{cases}
$$
To get a function that will work, we can use any smooth function $g(t)$ such that $g(t) = 0$ when $t<0$, $g(t)$ is increasing when $t \geq 0$, and $g(t) \to \infty$ as $t \to \infty$.  With such a function, we may define
$$
\phi(t) = (g(-t),g(t))
$$
note that this is exactly the same as defining
$$
\phi(t) = \begin{cases}
(g(-t),0) & t < 0\\
(0,g(t)) & t \geq 0
\end{cases}
$$

One example of such a $g$ is
$$
g(t) = \begin{cases}
0 & t \leq 0\\
t e^{-1/t} & t > 0
\end{cases}
$$
