$\binom{n}{k}$ counts the number of ways in which you can select $k$ items from a collection of $n$ items.
Generally speaking, when you multiply two counts, you are counting the number of ways in which the two selections can be carried out; when you add them, you are counting the number of ways in which one of two selections can be carried out. For instance, if you have 3 pants and 4 shirts, then you have $3\times 4$ ways of selecting what pair of pants and shirt to wear; if you have 3 routes you can take by car to get to school and 4 routes you can take by bike, then you have $3+4$ routes you can take to get to school by either car or bike.
The binomial coefficient on the right hand side, $\binom{n}{n-r}$, counts the number of ways in which you can select $n-r$ items from $n$ possibilities; note that this is the same as $\binom{n}{r}$, the number of ways of selecting $r$ items from $n$ possibilities: because selecting the $n-r$ that you want is equivalent to selecting the $r$ that you don't want.
So the right hand side counts the number of ways of selecting $r$ items from $n$ possibilities, $2^{n-r}$ times.
Let's look at the sum on the left hand side: the $k$th summand is $\binom{n}{k}\binom{k}{r}.$ The first factor counts the number of ways of selecting $k$ items from $n$ possibilities; the second factor counts the number of ways of selecting $r$ items from $k$ items.
One possible interpretation is: you have $n$ things; first you select $k$ of them, and then you select $r$ items from the $k$ you selected in the first place. That is, you can think of the factor $\binom{n}{k}$ as a first-step "culling" of the possibilities, and the second factor $\binom{k}{r}$ as the final selection process.
Note, however, that the number of ways of selecting $r$ times this way is not the same as the number of ways of selecting $r$ items. For instance, if $n=3$, $r=1$, and $k=2$, then you are trying to pick a single element from $\{1,2,3\}$; there are three ways of doing this (pick $1$, pick $2$, or pick $3$); but if you first pick two items and then you pick one, then there are two ways you can end up picking $1$ (first pick $1$ and $2$, then pick $1$; or first pick $1$ and $3$, then pick $1$), two ways you can end up picking $2$, and two ways you can end up picking $3$. So you end up with $6$ different ways, instead of $3$ possible outcomes...
Now, by adding all of these procedures from $k=r$ to $k=n$, what are you doing? You are counting the number of ways of taking a $2$-step process to select $r$ items: you could first select $r$ items, and then keep them ($k=r$); or you could first select $r+1$ items and then pick $r$ of them ($k=r+1$); or you can first select $r+2$ items, and then pick $r$ of them ($k=r+2$); ...; or you could just pick all $n$ items and then select $r$ of them ($k=n$).
So the left hand side is counting the total number of ways in which you can select $r$ items from $n$ possibilities in a $2$-step selecting procedure.
The right hand side is counting the number of ways of selecting $r$ items from $n$ possibilities, and then multiplying that by $2^{n-r}$.
That suggests trying to figure out whether the $2$-step selecting procedure will result in the same $r$ items being chosen exactly $2^{n-r}$ times.
In my example with $n=3$ and $r=1$, we have 1 way in which $1$ will be ultimately chosen if we first select just $1$ and then keep it ($k=r$); we have two ways in which $1$ will ultimately be chosen if we first select two items, which includes $1$, and then select $1$ ($k=r+1=2$, the case we figured out above); and there is one way in which we will ultimately choose $1$ if we first keep all three items and then select $1$. That's a total of $1+2+1= 4 = 2^{3-1}$ ways of ultimately ending with $1$, so the formula works out.
Try to see if you can show that you will always have exactly $2^{n-r}$ ways of ending with a particular selection of $r$ items through the $2$-step procedure. If you do, then that will establish the equality, because both sides of the equality are counting the same thing, in two different ways.