@LSpice has already proven my revised conjecture in the updated answer to https://mathoverflow.net/questions/377981/classification-of-not-necessarily-connected-compact-lie-groups?noredirect=1#comment959636_377981, but let me give another, closely related proof.

Since $1\to \mathrm{Inn}(G_0) \to \mathrm{Aut}(G_0) \to \mathrm{Out}(G_0) \to 1$ always splits, see https://mathoverflow.net/questions/378218/, we can choose a subgroup $R_0 \subseteq \mathrm{Aut}(G_0)$ for which the restriction of $\mathrm{Aut}(G_0) \to \mathrm{Out}(G_0)$ is an isomorphism. The inverse image of $R_0$ under the map $f:G \to \mathrm{Aut}(G_0)$ induced by conjugation is a subgroup $K \subseteq G$ whose intersection with $G_0$ is $Z(G_0)$.

 Multiplying any $g\in G$ by arbitrary $h \in G_0$ multiplies the associated $f(g) \in \mathrm{Aut}(G_0)$ by an arbitrary inner automorphism $f(h) \in \mathrm{Inn}(G_0)$, without changing $g$'s connected component. Thus, $K$ meets every connected component of $G$.

Using the result of https://mathoverflow.net/questions/150949, $K$ has a finite subgroup $R$ that meets every component of $K$, hence it meets every component of $G$ as well, and intersects $G_0$ within $Z(G_0)$. By design, the elements of $R$ either act by non-trivial outer automorphisms on $G_0$ or they act trivially on $G_0$. This proves my (revised) conjecture.

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COMMENT ADDED: An interesting, yet false, generalization is stated and disproven below.

It is well known that any compact, *connected* Lie group $G_0$ takes the form
$$G_0 = \frac{T^k \times G_1 \times \ldots \times G_\ell}{P}$$
where $T^k$ denotes a $k$-torus, $G_1, \ldots, G_\ell$ are compact, simply connected, simple Lie groups, and $P$ is central. One might think that the quotients in the expressions for $G$ and $G_0$ could be combined, so that any compact Lie group $G$ would take the form:
$$
G = \frac{(T^k \times G_1 \times \ldots \times G_\ell) \rtimes R}{P}
$$
where as before each element of $R$ acts by a non-trivial outer or acts trivially on $T^k \times G_1 \times \ldots \times G_\ell$. However, this is **false**.

Counterexample: Consider $G=(\mathrm{SO}(2k) \rtimes \mathbb{Z}_4) / \mathbb{Z}_2$, where the generator $r \in \mathbb{Z}_4$ acts by parity on $\mathrm{SO}(2k)$ and $r^2 = -1 \in SO(2k)$. Now let $G’=(\mathrm{Spin}(2k) \rtimes R)/P$ be a cover of $G$ whose connected component is $G_0'=\mathrm{Spin}(2k)$. There is some element $r'$ of $R$ that projects to $r$, hence $r’$ acts on $\mathrm{Spin}(2k)$ by parity. If $k$ is odd, then $Z(G_0') = \mathbb{Z}_4$, and $(r’)^2$ must be one of the two elements of order 4 in $Z(G_0')$ to project to $(r)^2 = -1$. However, parity exchanges these two elements, so we find $(r’)^{-1} (r’)^2 r’ \ne (r’)^2$, which is a contradiction. The case of even $k$ is very similar.