Does the Bezout GCD equation hold in a UFD? I'm wondering  when Bezout's Theorem is valid.
I know when the ring $R$ is a Euclidean Domain or a Principal Ideal Domain there is always $\gcd(a,b)$ and I can find $x,y \in R$ such that $\gcd(a,b) = ax + by.$

I know when the ring $R$ is a Unique Factorization Domain, there is always $\gcd(a,b)$. My question is, in this case can I find $x,y \in R$ such that $\gcd(a,b) = ax + by.$ ?

Now if $R$ is just a commutative ring, if there is $d \in R$ such that $\langle a\rangle+\langle b\rangle = \langle d\rangle $ I know that $d = \gcd(a,b)$. In this case, can I find $x,y \in R$ such that $\gcd(a,b) = ax + by$ ?
 A: The rings where it is valid Bézout's lemma are known as Bézout rings. In the case of integral domains rings we have the domains known as Bézout domains.
In general, a UFD doesn't have to be a Bézout domain. The classical example is $\Bbb{Z}[x]$, as Thomas Andrews pointed out in his comment. Theoretically there is a reason for which the above is true. We have the following:
Theorem 1: If $D$ is both a UFD and a Bézout domain, then $D$ is a PID.
Proof: See exercise 11 of section 8.3 of the book "Abstract Algebra" by Dummit and Foote.
By this theorem, if every UFD were also a Bézout domain, we would conclude that every UFD is a PID, and this is false (look at $\Bbb{Z}[x]$ again), as you surely know.
More generally, for Bézout domains we have the following result:
Theorem 2: Let $D$ be a Bézout domain. TFAE:
i) $D$ is a PID.
ii) $D$ is Noetherian.
iii) $D$ is a UFD.
iv) $D$ satisfies the ACCP (ascending chain condition on principal ideals).
v) $D$ is an atomic domain.
Proof: This is theorem 46 given in Pete L. Clark's notes on factorization in integral domains
Your second question has been also answered, but let me include a general version of what you wanted to prove.
Theorem 3: Let $a_1,a_2,\ldots a_n$ be nonzero elements of a commutative ring $R$. Then $a_1,a_2,\ldots a_n$ have a greatest common divisor $d$, expressible in the form $$d=r_1a_1+r_2a_2+\cdots +r_na_n$$ if only if the ideal $(a_1,a_2,\ldots a_n)$ is principal.
Note that $(a_1,a_2,\ldots a_n)$ is the same as $(a_1)+(a_2)+\cdots +(a_n)$.
Proof: This is theorem 6-3 in Burton's book "First Course in Rings and Ideals".
A: It is not necessarily true in a UFD.  For instance, if $k$ is a field, then the polynomial ring $k[s,t]$ is a UFD.  But $\gcd(s,t)=1$, and there do not exist $x$ and $y$ such that $sx+ty=1$.
For your second question, the answer is yes by definition.  Indeed, by definition, $\langle a\rangle+\langle b\rangle$ is the set of elements of the form $ax+by$.  So if $d\in \langle a\rangle+\langle b\rangle$, then there exist $x,y\in R$ such that $d=ax+by$.
The converse holds as well: if $d=\gcd(a,b)$ can be written in the form $ax+by$, then that means $d\in \langle a\rangle+\langle b\rangle$, and it follows that $\langle a\rangle+\langle b\rangle=\langle d\rangle$.
Thus a commutative ring has the property that the GCD of any two elements $a$ and $b$ can be written in the form $ax+by$ iff the sum of any two principal ideals is principal.  By induction on the number of generators, this is equivalent to any finitely generated ideal being principal.  Such a ring is known as a Bezout ring.  A Noetherian Bezout ring is the same thing as a principal ideal ring, but there exist non-Noetherian examples as well.  For instance, the ring of holomorphic functions on a connected open subset of $\mathbb{C}$ is a Bezout domain but has ideals which are not finitely generated.
A: There are simple examples of non-Bezout UFDs, e.g. any UFD of dimension $> 1$, e.g. $\Bbb Z[x,y],\,$ as follows by the below equivalent conditions for a UFD to be Bezout (or we can prove it directly: $\,\gcd(x,y)=1\,$ but $\,xf+y\,g = 1\Rightarrow 0 = 1,\,$ via eval at $\,x=0=y)$.
Theorem $\rm\ \  TFAE\ $ for a $\rm UFD\ D$ 
$(1)\ \ $ prime ideals are maximal if nonzero, $ $ i.e. $\rm\ dim\,\ D \le  1$
$(2)\ \ $ prime ideals are principal
$(3)\ \ $ maximal ideals are principal
$(4)\ \ \rm\ gcd(a,b) = 1\, \Rightarrow\, (a,b) = (1),\,$ i.e. $ $ coprime $\Rightarrow$ comaximal 
$(5)\ \ $ $\rm D$ is Bezout, i.e. all ideals $\,\rm (a,b)\,$ are principal.
$(6)\ \ $ $\rm D$ is a $\rm PID$
Proof $\ $ (sketch of $\,1 \Rightarrow 2 \Rightarrow 3 \Rightarrow 4 \Rightarrow 5 \Rightarrow 6 \Rightarrow 1)\ $ where $\rm\,p_i,\,P\,$ denote primes $\neq 0$
$(1\Rightarrow 2)$ $\rm\ \  p_1^{e_1}\cdots p_n^{e_n}\in P\,\Rightarrow\,$ some $\rm\,p_j\in P\,$ so $\rm\,P\supseteq (p_j)\, \Rightarrow\, P = (p_j)\:$ by dim $\le1$ 
$(2\Rightarrow 3)$ $ \ $ max ideals are prime, so principal by $(2)$
$(3\Rightarrow 4)$ $\  \rm \gcd(a,b)=1\,\Rightarrow\,(a,b) \subsetneq (p) $ for all max $\rm\,(p),\,$ so $\rm\ (a,b) = (1)$
$(4\Rightarrow 5)$ $\ \ \rm c = \gcd(a,b)\, \Rightarrow\, (a,b) = c\ (a/c,b/c) = (c)$ 
$(5\Rightarrow 6)$ $\  $ Ideals $\neq 0\,$ in Bezout UFDs are generated by an elt with least #prime factors
$(6\Rightarrow 1)$ $\ \ \rm (d) \supsetneq (p)$ properly $\rm\Rightarrow\,d\mid p\,$ properly $\rm\,\Rightarrow\,d\,$ unit $\,\rm\Rightarrow\,(d)=(1),\,$ so $\rm\,(p)\,$ is max

As for the gcd as the ideal sum in a PID, that follows using "contains = divides", viz.
$$ c\mid a,b\iff (c)\supseteq (a),(b)\iff (c)\supseteq (a)\!+\!(b)\!=\!(d)\iff c\mid d$$
Finally, yes: $\,d\in (a)\!+\!(b) = aR + bR $ $\iff d = ar + br'$ for some $\,r,r'\in R$
