Okay, now that I am reading a bit more of that page, I see that smoothness can be characterized in more concrete way:
Let $\phi: A \rightarrow B$ be a ring homomorphism making $B$ into a finitely presented $A$-algebra. We say that $\phi$ is standard smooth is $B$ can be realized in the form $B = A[x_1, ... , x_n]/(f_1, ... , f_c)$ for $c \leq n$, such that the image of the polynomial
$$\operatorname{det}\begin{pmatrix} \frac{\partial f_i}{\partial x_j}\end{pmatrix}_{1 \leq i, j \leq c} \in A[x_1, ... , x_n]$$
is a unit in $B$.
Then $f: X \rightarrow S$ is smooth if and only if for every $x \in X$, there exists an affine open neighborhood $U$ of $x$ and an affine open set $V$ of $S$ such that $f(U) \subset V$ and $U \rightarrow V$ corresponds to a standard smooth ring homomorphism. (Lemma 28.32.11).
Also, in the example I asked about, with $R$ a DVR and $X= \operatorname{Spec} A$ a scheme of finite type over $R$, (28.32.3) says that in order to say that $X$ is smooth over $R$, it is sufficient that $X$ be over $R$ and that the generic fibre $X \times_R \operatorname{Spec}K$ and special fibre $X \times_R \operatorname{Spec}k$ be smooth varieties in the usual sense.
Another definition of smoothness (given here): $f: X \rightarrow Y$ is smooth if it is locally of finite presentation, flat, and if for all $y \in Y$, $X \times_Y \operatorname{Spec} \kappa(y)$ is smooth as a scheme over the field $\kappa(y)$. Equivalently, this last condition can be restated as saying $X \times_Y \operatorname{Spec} \overline{\kappa(y)}$ is regular over $\overline{\kappa(y)}$.
I myself got an answer to this question.
First, note that the condition 2. is stable under the base change along the morphism between schemes of finite type over $k$. You need to prove this in order to get over Proposition 10.1(b).
Take an arbitrary $z\in Z$. Let $X_z$, and $Y_z$ be the fiber of $z$ in $X$, and $Y$. Take an arbitrary irreducible component $X'$ of $X_z$. We must show that the dimension of $X'$ is $m + n$.
The generic point of $X'$ goes into an irreducible component $Y'$ of $Y_z$, when $f(X')\subseteq Y'\Leftrightarrow X'\subseteq f^{-1}(Y')$. Then, $X'$ is an irreducible component of $f^{-1}(Y')$ also. Because of [1], $f^{-1}(Y') \to Y'$ can be viewed a base change of $f$, with $Y'$ endowed with the reduced closed subscheme structure, and $f^{-1}(Y')$ with some corresponding scheme structure. Since condition 2. is stable under base change in a nice situation, $f^{-1}(Y') \to Y'$ also satisfies condition 2., and so $\dim X' = \dim Y' + m = n + m$. You know the last equality through the condition 2. for $g$, and Corollary 9.6 of Hartshorne.
[1]Image of base change of immersion
Best Answer
$\Omega_{X/Y}\neq 0$. In particular, the module of differentials $\Omega_{(k[t]/(t^n)/k)}$, which is the global sections of $\Omega_{X/Y}$, is a $k[t]/(t^n)$ module with generator $dt$ and relations generated by $nt^{n-1}dt=0$. So $dt$ is nonzero when $n>1$.