You can obtain the $G=KAK$ decomposition from a decomposition of the type $F=UR$. To avoid unnecessary complications, let's assume that our reductive group $G$ is a selfadjoint subgroup of $\operatorname{GL}(n,\mathbb{R})$. Then the map $g \mapsto g^{-t}$ is an involution of $G$, which is called the Cartan involution and is typically denoted by $\theta$. The first observation to make is that the fixed-point set $K = \{ g \in G \colon \theta(g)=g \}$ of $\theta$ is a maximal compact subgroup of $G$. For example, if $G=\operatorname{GL}(n,\mathbb{R})$, then $K=\operatorname{O}(n)$.

Next we observe that $\theta$ induces an involution (also denoted by $\theta$) at the Lie algebra level: explicitly, this is the map $X \mapsto -X^t$. If $\mathfrak{p}$ denotes the $-1$-eigenspace of this latter involution, then one has the following result.

The map $K \times \mathfrak{p} \to G$ given by $(k, X) \mapsto k e^X$ is a diffeomorphism.

In particular, every $g \in G$ can be expressed as $k e^X$ for some $k \in K$ and $X \in \mathfrak{p}$. This decomposition is known as the Cartan decomposition; it is a generalization of the polar decomposition to $G$ (and is, I presume, the $F=UR$ decomposition stated in the OP). Indeed, if $G = \operatorname{GL}(n,\mathbb{R})$, then $\mathfrak{p}$ is just the set of symmetric matrices, and thus the set $\exp \mathfrak{p}$ consists of symmetric, positive semidefinite matrices.

Now let $\mathfrak{a}$ denote a maximal abelian subspace of $\mathfrak{p}$. Then it can be shown that $A = \exp \mathfrak{a}$ is a closed abelian subgroup of $G$ with Lie algebra $\mathfrak{a}$. It can also be shown that $\mathfrak{a}$ is unique up to conjugacy via an element of $K$. That is to say, if $\mathfrak{a}'$ is another maximal abelian subspace of $\mathfrak{p}$, then there is a $k \in K$ such that $\text{Ad}(k) \mathfrak{a} = \mathfrak{a}'$. With this information we can obtain the decomposition $G=KAK$: given $g \in G$, one observes that $p=gg^t \in \exp \mathfrak{p}$, say $p=e^X$. Thus there is a $k \in K$ such that $\text{Ad}(k)X \in \mathfrak{a}$, and then $e^{-\text{Ad}(k)X/2}kg \in K$ (because it is fixed by $\theta$), whence $g \in KAK$.

This hopefully alleviates your 3-terms-vs-2-terms issue.

I'm not aware of any relationship between the Iwasawa decomposition and the $KAK$ (polar) decomposition.