Regular elements need not be semisimple! For example, in the Lie algebra $\frak{sl}_2$, every non-zero element is regular, with the centralizer spanned by the element itself. Among the elements of the standard basis, $e$ and $f$ are nilpotent, whereas $h$ is semisimple. Only $h$ spans a Cartan subalgebra; the subalgebras spanned by $e$ and $f$ are normalized by $h$, so they are not self-normalizing.
Your statement that a regular element belongs to a Cartan subalgebra is false, due to a missing crucial hypothesis: the element needs to be semisimple. Indeed, in a semisimple Lie algebra, a Cartan subalgebra is a maximal abelian subalgebra consisting of semisimple elements.
$\DeclareMathOperator{\Tr}{Tr}
\DeclareMathOperator{\ad}{ad}
\DeclareMathOperator{\Lie}{Lie}
\newcommand{\g}{{\mathfrak g}}
\newcommand{\z}{{\mathfrak z}}
\newcommand{\s}{{\mathfrak s}}
\newcommand{\O}{{\mathcal O}}
\newcommand{\wh}{\widehat}
\newcommand{\wt}{\widetilde}$
Let $G$ be a connected compact Lie group, and let $\g$ denote its Lie algebra.
Write $\z$ for the center of $\g$ and set $\s=[\g,\g]$, which is a semisimple Lie algebra.
Then
$$ \g=\z\oplus\s.$$
Write $Z=Z(G)^0$ (the identity component of the center of $G$), $\ \wt S=[G,G]$, $S=G/Z(G)$.
Then $Z$ is a torus, whereas $\wt S$ and $S$ are semisimple Lie groups.
We can identify
$$ \z=\Lie Z,\quad \Lie S=\s=\Lie\wt S.$$
The group $G$ acts on $\g$ by adjoint representation.
Moreover, $G$ acts on $\g$ via the canonical surjective homomorphism
$\pi\colon G\to S$.
We write
$$ g\cdot X=s\cdot X,\quad\text{where} \ g\in G,\ s=\pi(g)\in S,\ X\in\g.$$
We write an element $X\in \g$ as
$$ X=X_\z+X_\s\ \quad \text{with}\ X_\z\in\z,\ X_\s\in \s\,.$$
Then
$$g\cdot X=X_\z+g\cdot X_\s=X_\z+s\cdot X_\s\,.$$
Let $\g^*$ denote the dual space for $\g$.
Then
$$\g^*=\z^*\oplus \s^*.$$
For $r\in \g^*$ we may write
$$r=r_\z+r_\s\quad\text{with}\ r_\z\in\z^*, \ r_\s\in \s^*.$$
Then for $g\in G$ we have
$$g\cdot r=r_\z+g\cdot r_\s=r_\z+s\cdot r_\s\,.$$
Let $X\in\g$, $\ X=X_\z+X_\s\,$. Then $[X_\z\,,X_\s]=0$.
It follows that
$$\exp -tX=(\exp -tX_\z)\cdot( \exp -tX_\s) \quad\text{with}\
\exp -tX_\z\in Z,\ \exp -tX_\s\in\wt S,$$
whence
$$(\exp -tX)\cdot \alpha=\alpha_\z+(\exp-t X_\s)\cdot \alpha_\s\quad
\text{for}\ \alpha=\alpha_\z+\alpha_\s\in \g^*.$$
Write
$$ \wh X=\frac d{dt}\Big|_{t=0}(\exp -tX)\cdot \alpha.$$
Then $\wh X=\wh{X_\s}$, where
$$ \wh {X_\s}=\frac d{dt}\Big|_{t=0}(\exp -tX_\s)\cdot \alpha_\s.$$
For $r=r_\z+r_\z \in\g^*$ write $\O_r=G\cdot r$, $\ \O_{r_\s}=S\cdot r_\s\,$. Then
$$ \O_r=r_\z+S\cdot r_\s=r_\z+\O_{r_\s}\,.$$
Thus for $\alpha=g\cdot r\in \O_r$ we have
\begin{equation}\label{e:*}
T_\alpha(\O_r)\cong T_{\alpha_\s}(\O_{r_\s}).\tag{$*$}
\end{equation}
Consider the adjoint representation
$$\ad\colon \g\to\mathfrak{gl}(\g),\quad (\ad X)\cdot Y=[X,Y].$$
For $X=X_\z+X_\s\in\g$ we have $\ad X=\ad X_\s\,$.
Consider the Killing form
$$ k\colon \g\times\g\to{\mathbb R}, \quad
(X,Y)\mapsto \Tr\!\big ((\ad X)\cdot (\ad Y)\big).$$
Then
$$ k(X,Y)=\Tr\!\big ((\ad X_\s)\cdot (\ad Y_\s)\big)=k(X_\s\,,Y_\s).$$
For $\lambda \in\g$, we define a skew-symmetric form on $T_\alpha(\O_r)$ by
$$\omega_\lambda(\wh X,\wh Y)=-k(\lambda,[X,Y]).$$
Since $[X,Y]=[X_\s\,,Y_\s]$, for $\lambda=\lambda_\z+\lambda_\s$ we have
$$\omega_\lambda(\wh X,\wh Y)=-k(\lambda,[X,Y])
=-k(\lambda_\s,[X_\s\,,Y_\s])=\omega_{\lambda_\s}(\wh{X_\s}\,, \wh{ Y_\s\, }\,).$$
We can identify the tangent spaces
$T_\alpha(\O_r)$ and $T_{\alpha_\s}(\O_{r_\s})$ by \eqref{e:*}.
Then the formula above is probably what you need.
Best Answer
It is possible to have a non-trivial, closed, connected Lie subgroup $K$ of a compact, connected Lie group $G$ such that no element of $\mathfrak k$ is regular in $\mathfrak g$. For example, consider $K = \operatorname{SU}_2$ embedded in $G = \operatorname{SU}_4$ as $\begin{pmatrix} a & b \\ c & d \end{pmatrix} \mapsto \begin{pmatrix} a &&& b \\ & a & b \\ & c & d \\ c &&& d \end{pmatrix}$.
On second thought, even this is overkill; you could just take $K$ to be the (non-trivial) central torus in $G = \operatorname U_2$. But maybe you wanted a non-central as well as non-trivial example.