In Eisenbud and Harris' 3264 and all that, part of their statement of Kleiman's theorem is the following:
Suppose that an algebraic group $G$ acts transitively on a variety $X$ over an algebraically closed field of characteristic $0$, $A \subseteq X$ a subvariety. Then
(a) If $B \subset X$ is another subvariety, there is an open dense set of $g \in G$ s.t $gA$ is generically transverse to $B$.
(b) If $\varphi:Y \rightarrow X$ is a morphism of varieties, then for general $g \in G$ the preimage $\varphi^{-1}(gA)$ is generically reduced and of the same codimension as $A$.
The proof of Kleiman's theorem states that part (a) follows from part (b), with $Y = B$ (and presumably $\varphi = \iota$, the inclusion map). As far as I can tell, part (b) just implies that $B \cap gA$ is generically reduced and $\operatorname{codim}_B(gA \cap B) = \operatorname{codim}_X(A)$ for an open dense subset of $g \in G$. I don't see how generic transversality follows from this.
Claim: If $Z_1$ and $Z_2$ intersect properly at $x\in X$ but not transversely, then $Z_1\cap Z_2$ is singular at $x$.
Proof: The problem is local, so we may assume $X$ is affine and embed it in $\Bbb A^n$. Suppose $I_1,I_2$ are the ideals cutting out $Z_1,Z_2$ respectively. Then the scheme-theoretic intersection $Z_1\cap Z_2$ is cut out by $I_1+I_2$, so the Jacobian matrix of $Z_1\cap Z_2$ at $x$ is the concatenation of the Jacobian matrices for $Z_1$ and $Z_2$ at $x$. As the tangent space is the kernel of the Jacobian, this gives that $T_x(Z_1\cap Z_2)= T_xZ_1\cap T_xZ_2$ inside $T_xX$.
As $T_xZ_1+T_xZ_2\subsetneq T_xX$, we have that $\dim T_xZ_1 \cap T_xZ_2 > \dim T_xZ_1+\dim T_xZ_2 -\dim T_xX$. Since $X$ is smooth (from the assumption it admits a transitive group action), $\dim T_xX=\dim X$, while we know from basic properties of the tangent space that $\dim T_xZ_1 \geq \dim_x Z_1$, so $$\dim T_xZ_1+\dim T_xZ_2 -\dim T_xX \geq \dim_x Z_1 +\dim_x Z_2-\dim X.$$ But the RHS is the dimension of $Z_1\cap Z_2$ at $x$, so the dimension of the tangent space to $Z_1\cap Z_2$ at $x$ is strictly larger than the dimension of $Z_1\cap Z_2$ at $x$, or $Z_1\cap Z_2$ is singular at $x$. $\blacksquare$
In particular, this implies that if $Z_1$ and $Z_2$ intersect properly at a point and $Z_1\cap Z_2$ is smooth there, then $Z_1$ intersects $Z_2$ transversely there. The condition of intersecting properly is taken care of by the phrase "for a general $g\in G$, the preimage is ... of the same codimension...". Now since a generically reduced scheme of finite type over a perfect field is generically smooth, we're done.