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.
You have a proof which uses the fact that maximal compact open subgroups correspond to vertices in the building of ${\rm GL}(n)$. For instance for the Cartan decomposition you want to classify the orbits of $K={\rm GL}(n,{\mathfrak o}_F)$ (${\mathfrak o}_F$ denotes the ring of integers of your p-adic field $F$) in the vertex set of the building. To the aim, you first use the fact that $K$ acts transitively on the apartments containing the vertex fixed by $K$. This way you may send any pair of vertices $(s,t)$ in a fixed apartment $A$. The stabilizer of $s$ acts transitively on the Weyl chambers and you may assume that $t$ lies in a fixed Weyl chamber. From this it's easy to prove that
$$
t={\rm Diag}(\varpi_F^{k_1},...,\varpi_F^{k_n}).s
$$
where e.g. $k_1 \geq k_2 \geq ...\geq k_n$ and where $\varpi_F$ is a uniformizer of $F$.
Similarly the Iwasawa decomposition $G=KB$ corresponds to the fact that $K$ acts transitively on the germs of "quartiers" (you may prove this using the fondamental properties of the building stated in Bruhat-Tits, IHES, volume 1).
In fact doing the proofs this way is cheating ! For Bruhat and Tits prove the basic properties of the building by using properties of the valuated root datum, that is by doing row and column operations on matrices ...
But even if these proofs are not geniune "proofs", they allow you to "visualize" why these decompositions hold.
Best Answer
About the question of uniqueness of maximal compact subgroups, up to conjugation, I propose a Riemannian geometric approach:
Theorem (É. Cartan). A compact group of isometries of a nonpositively curved complete simply-connected Riemannian manifold has a fixed point.
Proof. (Rough sketch) Consider any orbit. It is compact. By convexity of distance functions (the curvature of the ambient is nonpositive), its center of mass can be defined and it is plainly a fixed point. QED
Now one uses the fact that the symmetric space of non-compact type $G/K$ is nonpositively curved. If $H$ is a compact subgroup of $G$ then it has a fixed point $gK$ by the theorem, so $g^{-1}Hg$ fixes the basepoint $1K$ and hence is contained in $K$.
Edit: on suggestion of Ben, I complete my answer as follows. Let $\mathfrak g = \mathfrak k + \mathfrak p$ be the decomposition of the Lie algebra of $G$ into the eigenspaces of the involution. Since $M$ is complete and nonpositively curved, the Hadamard-Cartan theorem says that the map $\varphi:\mathfrak p \to G/K$ given by $\varphi(X)=(\exp X)K$ is a smooth covering. Next one sees that $\varphi$ is injective (assume $\varphi(X)=\varphi(Y)$, apply $\mathrm{Ad}$ to both sides and use the uniqueness of polar decomposition of a matrix, and the semisimplicity of $\mathfrak g$, to deduce that $X=Y$). Hence there are diffeomorphisms $\mathbf R^n\cong\mathfrak p\cong G/K$. Now it follows rather easily that $\mathfrak p\times K\to G$, $(X,k)\mapsto(\exp X)k$ is a diffeomorhism ($G=\exp[\mathfrak p]\cdot K$ in general for a symmetric space).