In Eisenbud-Harris' 3264 and All That, on page 33 they state: Subschemes $Y$ and $Z$ of $X$ have generic transversality iff dimensional transversality and each connected component of $Y\cap Z$ has a smooth point at which intersection is reduced. Then they state: smoothness of $X$ is necessary, since inside of $k[s^2, st, t^2]$, the ideal $(s^2)$ defines a doubled-line which intersects the vertex of $\text{Spec }k[s^2,st,t^2]$ at a reduced point of the reduced line given by $(st,t^2)$. Here, $p=(0,0,0)$ is a closed point.
Dimensional transversality here means $codim_X(Y\cap Z) = codim_X(Y)+codim_X(Z)$.
Generic transversality means that the span of the tangent spaces at $p \in Y\cap Z$ add: $T_p Y + T_p Z = T_p X$.
I am failing to see how generic transversality does not hold. I gather that $T_pX$ is not necessarily the Zariski tangent space (defined as $Hom_{k(p)}(\text{Spec }k[\epsilon]/\epsilon^2, X)$ where $\mathfrak{p}\mapsto (\epsilon)$), since those do add to $T_pX$ in this case. Rather, $T_pX$ is to be taken as the affine tangent space defined by linear forms in partial derivatives of defining equations evaluated at $p$. Up to isomorphism, I take $X = \text{Spec } k[x_0,x_1,x_2]/<x_0x_2-x_1^2>$, $Y=V(x_0) \subset X$, $Z = V(x_1,x_2)\subset X$. The affine tangent spaces are:
$T_p X = V(0) = \mathbb{A}^3$
$T_p Y = V(x_0)$ gives the $x_1x_2$-plane in $\mathbb{A}^3$
$T_p Z = V(x_1,x_2)$ gives the $x_0$-axis.
I fail to see how $T_p Y + T_p Z \neq T_pX$.
Edit/Update: Brighter minds than me suggest that possibly the intended counterexample takes $\tilde{Y}=V(s^2,st)$ rather than $Y=V(s^2)$ so that we have dimensional transversality, $\tilde{Y}$ and $Z$ intersect in a reduced point at which $X$ is not smooth and $\tilde{Y}$ and $Z$ are not generically transverse. So it's possible that this was the intended counterexample/illustration of the importance of $X$ being smooth.
This leaves open the question (modulo my misunderstanding) of whether the original setup above provides a counterexample to the theorem.
Edit/Update 2: I believe that generic transversality here fails due to convention: $X$ must be smooth at $p$ in order to say that $Y$ and $Z$ are generically transverse there. So we have all conditions from the ``right-hand side" above (generic transversality, $Y$ and $Z$ intersect in a reduced point) except that $X$ is not smooth at $p$, and generic transversality fails.