How do you make a joint distribution table of $(Y_1,Y_2)$?? If you let $X_1$ and $X_2$ be the numbers obtainted on two rolls of a fair die. Let $Y_1 = max(X_1 , X_2), Y_2 = min(X_1 , X_2)$.
First of all how do I interpret $Y_1$ and $Y_2$, and then second, how do I form a joint probability table from this?

Here $X_1$ is intended to the the number obtained on the first toss, and $X_2$ the result on the second. Let random variable $Y=\min(X_1,X_2)$ denote the minimum of $X_1$ and $X_2$, and $Z=\max(X_1,X_2)$ the maximum. 
So for example if we had a $5$ on the first toss, and $2$ on the second, then $Y=2$ and $Z=5$.  If we had a $3$ on each, then $Y=Z=3$.  
To express it in game terms, the two dice are tossed, and Alicia gets, in dollars, the smaller of the two numbers showing. Then random variable $Y$ represents Alicia's gain. 
We want to draw up a $6\times 6$ table of probabilities that tells us the probability that $Y=a$ and $Z=b$ for all pairs $a,b$.  Kind of like a mileage chart. 
From the definition, we have $Y\le Z$. So for any pair $(a,b)$ such that $a\gt b$, we have probability that $Y=a$ and $Z=b$ is $0$. So in your table, write $0$ for all such situations. The table is $6\times 6$, and we have already filled in $15$ places. Only $21$ to go.
For all pairs $(a,b)$ with $a\lt b$, we can get $Y=a$ and $Z=b$ in two ways: our first toss was $a$ and the second was $b$, or the other way around.
So if $a\lt b$, then with probability $\dfrac{2}{36}$, we have $Y=a$ and $Z=b$. This fills in $15$ more entries. You may prefer to enter them as $\dfrac{1}{18}$. 
Finally, we want to find the entries on the main diagonal. What is, for example, the probability that $Y=4$ and $Z=4$? The $\max$ and $\min$ are both $4$  precisely if both tosses result in a $4$. The probability of this is $\dfrac{1}{36}$. All the diagonal entries, by the same reasoning, have value $\dfrac{1}{36}$.
As a check, let's see whether the table entries add up to $1$. There were $15$ entries of $\dfrac{2}{36}$, and $6$ of $\dfrac{1}{36}$, so the total is
$15\cdot\dfrac{2}{36}+6\cdot\dfrac{1}{36}$.  This  sum is indeed $1$. 
