All numbers are integers in the following unless stated otherwise.
Answer to your first question:
Assume that $\gcd(c_1,c_2)= a$ then by Bézout´s identity we have that there exists $m,n\in\mathbb{Z}$ such that
$$
a=n\cdot c_1+ m\cdot c_2,
$$
and the fact that $\frac{d}{a}$ leaves $0$ as a remainder can be phrased as
$$
d=k\cdot a,\qquad k\in\mathbb{Z}.
$$
Putting these two together means
$$
d= k\cdot n\cdot c_1+k\cdot m\cdot c_2
$$
to make this a bit cleaner let $\tilde{n}=k\cdot n$ and $\tilde{m}=k\cdot m$ so
$$
d=\tilde{n}\cdot c_1+\tilde{m}\cdot c_2, \qquad \tilde{n},\tilde{m}\in\mathbb{Z}
$$
So we could get $d$ as a linear combination of $c_1$ and $c_2$, translate this as we could get $d$ by taking a number of times $c_1$ and a number of times $c_2$. Observe that these numbers can be negatives as well so put a couple of $c_1$s in and remove some $c_2$s is also possible.
Now for the second question:
Assume that $a,b\in\mathbb{Z}$. It is not hard to see that then
$$
b>0 \ \ \text{and} \ \ \ b\mid a\Rightarrow \gcd(a,b)=b
$$
Another ingredient is the following fact, if $b\neq 0$ then
$$
a=b\cdot q+r\Rightarrow \gcd(a,b)=\gcd(b,r)
$$
where $b>r$. I chellenge you to try to prove this :)
Having these in mind the Eucledean algorithm is as follows:
$$
a=bq_1+r_1 \qquad 0\le r_1< b
$$
$$
b=r_1q_2+r_2 \qquad 0\le r_2< r_1
$$
$$
r_1=bq_3+r_2 \qquad 0\le r_2< r_3
$$
and so on. This will terminate after a while since the remainders are getting smaller and smaller and at some point you will reach $r_{n+1}=0$.
This gives us the following chain of equalities:
$$
\gcd(a,b)=\gcd(b,r_1)=\gcd(r_1,r_2)=\ldots=\gcd(r_{n-1},r_n)=\gcd(r_n,0)=r_n
$$
I hope this helped a bit.
Best Answer
If you are interested in counting tuples $(a_1,a_2,\dots,a_n)$ such that $\gcd(a_1,\dots,a_n) = G$ and $\operatorname{lcm}(a_1,\dots,a_n) = L$ then we can do it as follows.
If $L/G = \prod\limits_{i=1}^s p_i^{x_i}$ then each $a_i$ must be of the form $G \prod\limits_{j=1}^s p_i^{y_{i,j}}$ with $0 \leq y_{i,j} \leq x_i$.
Hence for each prime $p_i$ we require that the function from $\{1,\dots, n\}$ to $\mathbb N$ that sends $j$ to $y_{i,j}$ be a function that hits $0$ and $x_i$.
The number of such functions is easy by inclusion-exclusion for $x_i \geq 1$, it is $(x_i+1)^n - 2(x_i)^n + (x_i-1)^n$.
It follows the total number of tuples is $\prod\limits_{i=1}^s ( (x_i+1)^n - 2x_i^n + (x_i-1)^n)$.