Given big rectangle of size $x, y$, count sum of areas of smaller rectangles.

Let's say we have two integers $$x$$ and $$y$$ that describe one rectangle, if this rectangle is splitten in exactly $$x\cdot y$$ squares, each of size $$1\cdot 1$$, count the sum of areas of all rectangles that can be formed from those squares.

For example if $$x = 2, y = 2$$, we have $$4$$ rectangles of size $$(1,1)$$, $$2$$ rectangles of size $$(1, 2)$$, $$2$$ rectangles of size $$(2, 1)$$ and $$1$$ rectangle of size $$(4,4)$$. So the total answer is $$4\cdot 1+ 2\cdot2 + 2\cdot2+1\cdot4 = 16$$

Is there easy way to solve this for different $$x$$ and $$y$$?

I assume, in your example, you mean there is one rectangle of size $$(2, 2)$$, because there are none $$(4, 4)$$.

In general, we have $$(x - a + 1) \times (y - b + 1)$$ rectangles of size $$(a, b)$$: we need to choose how many empty columns we leave before the rectangle $$(0, 1, 2, \dots \text{ or } x-a)$$, and how many rows we leave above it $$(0, 1, 2, \dots \text{ or } y-b)$$

So, in total, we have this area $$\sum_{a=1}^x \sum_{b=1}^y (x - a + 1) \times (y - b + 1) \times ab = \\ \sum_{a=1}^x \Big( a (x - a + 1) \sum_{b=1}^y b (y - b + 1)\Big) = \\ \big(\sum_{a=1}^x a (x - a + 1)\big) \big( \sum_{b=1}^y b (y - b + 1)\big)$$

We can simplify this further by examining the sequence $$A_n = \sum_{a=1}^n a (n - a + 1)$$.

If we examine the first few elements of this sequence $$(1, 4, 10, 20, 35, 56)$$, we can notice that the sequence of differences $$(1, 3, 6, 10, 15, 21)$$ is exactly the sequence of triangular numbers. If that holds true in general, $$A_n$$ is the sum of the first $$n$$ triangular numbers, i.e. $$A_n = \frac{n(n+1)(n+2)}{6}$$ as seen here.

Here's a proof that this indeed holds in general: \begin{align} A_n - A_{n-1} = 1 \times n &+ 2 \times (n-1) + 3 \times (n-2) + \dots + n \times 1 \\ &- 1 \times (n-1) - 2 \times (n-2) - 3 \times (n-3) - \dots = (n-1) \times 1 = \\ &= n + (2 - 1) \times (n - 1) + (3 - 2) \times (n - 2) + \dots + (n - 2 - n + 1) \times 1 \\ &=\sum_{i=1}^n i = \frac{n(n+1)}{2}, \end{align} which is indeed the $$n^{\text{th}}$$ triangle number.

Thus, the answer to the original question is $$\frac{x(x+1)(x+2)}{6} \times \frac{y(y+1)(y+2)}{6}$$ and, in the case $$x=y=2$$ this is indeed 4.

It is useful to move from simpler cases to more complex ones!

Consider $$1\times y$$ rectangle. The sum of areas of all rectangles is: \begin{align}A_{1\times y}&=1\times y+2\times (y-1)+3\times (y-3)+\cdots +y\times 1=\\ &=\sum_{i=1}^yi\cdot (y-i+1)=\\ &=(y+1)\sum_{i=1}^y i-\sum_{i=1}^yi^2=\\ &=(y+1)\cdot \frac{y(y+1)}{2}-\frac{y(y+1)(2y+1)}{6}=\\ &=\frac{y(y+1)[3y+3-2y-1]}{6}=\\ &=\frac{y(y+1)(y+2)}{6}.\end{align}

Consider $$2\times y$$ rectangle. The sum of areas of all rectangles is: \begin{align}A_{2\times y}&=\color{red}{2}\times [1\times y+2\times (y-1)+3\times (y-3)+\cdots +y\times 1]+\\ &+\color{red}{1}\times [2\times y+4\times (y-1)+6\times (y-3)+\cdots +2y\times 1]=\\ &=\color{red}{2}\times [A_{1\times y}]+\color{red}{1}\times 2\times [A_{1\times y}]\\ &=(\color{red}2+\color{red}1\times 2)\times [A_{1\times y}]\\ &=4\cdot \frac{y(y+1)(y+2)}{6}.\end{align}

Consider $$3\times y$$ rectangle. The sum of areas of all rectangles is: \begin{align}A_{3\times y}&=\color{red}{3}\times [1\times y+2\times (y-1)+3\times (y-3)+\cdots +y\times 1]+\\ &+\color{red}{2}\times [2\times y+4\times (y-1)+6\times (y-3)+\cdots +2y\times 1]+\\ &+\color{red}{1}\times [3\times y+6\times (y-1)+9\times (y-3)+\cdots +3y\times 1]=\\ &=(\color{red}{3}+\color{red}2\times 2+\color{red}1\times 3)\times [A_{1\times y}]=\\ &=10\cdot \frac{y(y+1)(y+2)}{6}.\end{align}

And now consider $$x\times y$$ rectangle. The sum of areas of all rectangles is: \begin{align}A_{x\times y}&=\ \ \ \ \ \ \ \ \ \ \color{red}{x}\times [1\times y+2\times (y-1)+3\times (y-3)+\cdots +y\times 1]+\\ &+\color{red}{(x-1)}\times [2\times y+4\times (y-1)+6\times (y-3)+\cdots +2y\times 1]+\\ &+\color{red}{(x-2)}\times [3\times y+6\times (y-1)+9\times (y-3)+\cdots +3y\times 1]=\\ & \ \ \vdots\\ &+\ \ \ \ \ \ \ \ \ \ \ \color{red}{1}\times [x\times y+2x\times (y-1)+3x\times (y-3)+\cdots +xy\times 1]=\\ &=(\color{red}{x}+\color{red}{(x-1)}\times 2+\color{red}{(x-2)}\times 3+\cdots +\color{red}1\times x)\times [A_{1\times y}]=\\ &=[A_{x\times 1}]\cdot [A_{1\times y}]=\\ &=\frac{x(x+1)(x+2)}{6}\cdot \frac{y(y+1)(y+2)}{6},\end{align} which complies with the brilliant answer given by Todor Markov (+1).