Consider the case of a group $K$ acting on $X$. Restricting to the orbit of the point $x$, by the orbit-stabilizer theorem $X$ is identified with $K/\text{Stab}(x)$. Your question seems to amount to asking whether the projection $K\times K/\text{Stab}(x)\to K/\text{Stab}(x)\times K/\text{Stab}(x)$ given by $(\gamma,p)\mapsto (\gamma\cdot p,p)$ admits an immersed section. Restricting such a section to a fiber $K/\text{Stab}(x)\times \text{pt}$ would give an section of the map $K\to K/\text{Stab}(x)$, which will not exist in general. (Edit: rewritten. Thanks to David Carchedi for pointing out the mistake in my notation.)

A concrete counterexample: consider the action of $K=\mathbb{R}$ on $X=S^1$ by translation. Then $G$ is diffeomorphic to $\mathbb{R}\times S^1$, and $H$ would have to be diffeomorphic to $S^1\times S^1$, but $S^1\times S^1$ does not immerse in $\mathbb{R}\times S^1$. Moreover the restriction to a fiber $S^1\times \text{pt}$ would be an immersion of $S^1$ into $\mathbb{R}$, which also cannot exist.

But this is not a phenomenon just of fundamental group or of equi-dimensional immersions. For example, consider $K=\text{Isom}(\mathbb{H}^2)=\text{PSL}_2\mathbb{R}$ acting on a hyperbolic surface $\Sigma$. A section of $\text{PSL}_2\mathbb{R}\times \Sigma\to \Sigma\times \Sigma$ restricts to an immersed section of $\text{PSL}_2\mathbb{R}\to \Sigma$. Of course $\Sigma$ *does* immerse, in fact embed, in $\text{PSL}_2\mathbb{R}$ (it's a 3-manifold). But if we had a continuous section $\varphi:\Sigma\to \text{PSL}_2\mathbb{R}$, we could translate a nonzero vector $v$ to each point $p$ by the isometry $\varphi(p)$. This would yield a nonzero vector field on $\Sigma$, contradicting the Gauss-Bonnet theorem. The same works for $\text{SO}(3)$ acting on $S^2$ or $\mathbb{R}P^2$, where the fundamental group is not the issue.