Let's take a smaller example to understand our problem. Let n=3 and k=2.
Let's say we want to know first how n=2 numbers can be partitioned into k=2 subsets. Let's say it is x. Now what we can do is add 3 to any subset in each of the x partitions. That way we will now have n=3 numbers partitioned into 2 subsets.
Also, we can do it another way. Now we ask how to partition n=2 numbers but in k=1 subsets. Let's say it is y. Now we can just make 3 separate sets and add them to those y partitions. Since all the y partitions had k-1 sets adding 3 separately will make it k.
Clearly, this is very similar to the previous problem. Now, what are the total count = 2*x + y. In a more generic manner if f(n, k) is the count of the partitions of n into k subsets then we can represent it as: f(n, k) = (k)*f(n-1, k) + f(n-1, k-1)

If you had any difficulty understanding this part watch the video .

Now no matter what n is if we have to partition it into 1 subset there is just 1 way hence for all n, k=1 the value will 1 too.

Another situation is when k>n, we literally can never make k non-empty partitions with n numbers. So those will be 0. Also if n==k then there is just one way to implement it. Each subset with 1 number.
But in all other cases, we will use our recursive formula. So let's do it, dp[2][1] Clearly, k>n so 0

Now dp[2][2] Clearly k==n so 1.

Now dp[2][3]. Here n>k. So we know dp[2][3] = 2*dp[2][2] + dp[1][2] = 2*1 + 1 = 3

Now for dp[2][4] Again n>k So we use dp[2][4] = 2*dp[2][3] + d[1][3] = 2*3 + 1 = 7

I would insist you complete the grid yourself. If you are stuck watch the video.