I don't know how Bruhat-Tits theory is used in representation theory, but I think there is a little confusion of notions in your question.

For a connected reductive algebraic group $G$, given a point $x$ in $B(G,K)$, Bruhat and Tits define $4$ integral models denoted $\mathfrak{G}_x^0$, $\mathfrak{G}_x$, $\hat{\mathfrak{G}}_x$ and $\mathfrak{G}_x^{\dagger}$. Here $\mathfrak{G}_x^0$ is the connected component of $\mathfrak{G}_x$, and in Bruhat-Tits terminology $\mathfrak{G}_x^0(\mathcal{O}_K)$ is called the "connected fixator" ("fixateur connexe" in French). I guess that's the group you are referring to as "the connected component of the stabiliser of $x$" (I put that in quote, because to me this doesn't mean anything: when compact, the stabiliser of $x$ in $G(K)$ is a profinite group, i.e. it has normal closed (for the strong topology) subgroups of finite index, and the index can be as big as you want).

Ok, now if we look at $GL_2$, take a point $x$ in $B(G,K)$ and an apartment $A$ containing it. Then throw in many basis as in Bruhat-Tits I, 10.2, so that x identifies with $0\in \mathbf{R} = A$. Let $S_{-1}$ be the stabiliser of $-1\in A$. By Bruhat-Tits I, $10.2.9$, $S_{-1}$ consists of matrices $g\in GL_2(K)$ such that the valuation of the entries are greater then $\begin{pmatrix} 0&1\\-1&0 \end{pmatrix}+\frac{\omega(\textrm{det}(g))}{2}$. For example, $\begin{pmatrix} 0&\pi \\\pi^{-1}&0 \end{pmatrix}$ belongs to $S_{-1}$.

On the other hand, by Bruhat-Tits II, $4.6.28$, matrices in $\mathfrak{G}_{-1}^0(\mathcal{O}_K)$ have determinant in $\mathcal{O}_K^{\times }$ and belongs to $S_{-1}$ as well. So we conclude that the element $g = \begin{pmatrix} 0&\pi^2\\1&0 \end{pmatrix}$ belongs to the stabiliser of $-1$, but not to $Z(G(K))\mathfrak{G}_{-1}^0(\mathcal{O}_K)$.

So this is a negative answer to your first question, even if we interpret "connected component of the stabiliser of x" to mean "stabilisers of $x$ acting type preservingly".

EDIT: Ok, as pointed by Watson Ladd, the above counterexample is just not one, sorry that was silly. At least there is still the counterexample if you take $x=\frac{1}{2}$. Then I believe $g = \begin{pmatrix} 0&1\\\pi&0 \end{pmatrix}$ stabilises $\frac{1}{2}$, but does not belong to $Z(G(K))\mathfrak{G}_{\frac{1}{2}}^0(\mathcal{O}_K)$. Here the action of $g$ is not type preserving, though, so this calls for the question whether $Z(G(K))\mathfrak{G}_{x}^0(\mathcal{O}_K)$ consist of stabilizers of $x$ acting type preservingly.

abstractgroup $Z(G(K))P_x(O_K)$ is the identity component of a group scheme. $\endgroup$wouldencourage you to say "reductive butnon-semisimple groups" in your title; I clicked expecting a question on unipotent groups.) $\endgroup$