I came across two different definitions of regular point of a smooth map.
Let $X$,$Y$ be two smooth manifolds and let $f: X \rightarrow Y$ be a smooth map. Write $$ \mbox{rank} \; df = \max\{ \mbox{rank}\; d f(x): x \in X \}. $$
Definition 1: We say $x \in X$ is a regular point if $\mbox{rank}\; d f(x) = \mbox{rank}\; df$. Otherwise it is a critical point.
Definition 2: We say $x \in X$ is a regular point if the differential $ df(x): T_xX \rightarrow T_{f(x)} Y$ at $x$ is surjective. Otherwise it is a critical point.
Now suppose $Y$ is simply $\mathbb{R}^n$. Suppose $x$ is a critical point using definition 2. Then I can simply add one more dimension to $Y$. Define $f_1:= [f;f_0]$ where $f_0(x) = 0 \; \forall x \in X$. So $f_1$ is a smooth map from $X$ to $\mathbb{R}^{n+1}$. Now $x$ is a regular point by Definition 1 but a critical point by Definition 2 since the tangent space of $\mathbb{R}^{n+1}$ is just $\mathbb{R}^{n+1}$. So I am confused...
My guess is that there must be a way to reduce $Y$ to a "minimal dimension" so this scenario doesn't happen?
$\operatorname{rank} df$ is the maximal rank of all $df(x)$. I am not sure what its purpose should be. Obviously we have $\operatorname{rank} df \le \min (\dim X, \dim Y)$, but there is no reason why $\operatorname{rank} df = \dim Y$. Actually this is impossible if $\dim X < \dim Y$. But even if $\dim X \ge \dim Y$ we may have $\operatorname{rank} df < \dim Y$; $\operatorname{rank} df$ may take may take any value between $0$ and $\dim Y$.
Thus definition 1. does not seem to produce something useful since any smooth map would by definition have regular points. You should ask the person who gave the definition what has been his intention.
Anyway, each regular point in the sense of 2. is one in the sense of 1. (because in that case $\operatorname{rank} df = \dim Y$). The converse fails unless $\operatorname{rank} df = \dim Y$ (and this is in general not true).