As with most things in areas of mathematics and the sciences it is more the notation that is difficult to understand than the actual work, at least at this level. I'll explain what he did line by line.
1- He approached this problem by first thinking "what makes the ball move to begin with? What did the ball start with? What will cause the motion of the ball change?" So if I say "A ball is thrown" than I instantly know two things, first that the ball is moving forward and that the ball is moving above the earth. So he started with the fact that the ball is above the ground. So gravity will be the force acting on the ball to pull it down. So F=ma than when he says (d^2y/dt^2), that is the exact same thing as writing "a." (d^2y/dt^2) just means that he took the second derivative of a position function, a mathematical way to determin the postition of an object based on time (such as the equation y=3x, where x is some measure of time (i.e. 5seconds), which is the acceleration of an object. So since gravity always acts 'down' than "a" is negative and the acceleration on the surfface of the earth is always a constant "g" or 9.81m/s^2.
2- For the next line he says that he integrated F=-mg to obtain the next equation. While that is the true way to obtain the next equation, he more than likely had this memorized or looked on a reference table availible to anyone pretty much through all of highschool and a good bit of college. The equation is a basic kinematic equation, an equation that describes the motion of an object in a single direction. the first part of the equation y(0) just means that you must first factor in what height you originally threw the ball from, (say 1.5m or so). The next term factors in any velocity that the ball originally has, how quickly you throw the ball, (10m/s or so). The last term factors in the gravity from the previous line. Gravity will slow the initial velocity of the ball while the ball is travelling up, and will speed up the velocity of the ball after it reaches its highest point and is on the way back down. Useing this equation you can determin the position of the ball at any point vertically based on the amount of time that has passed. (note that this equation has its variable "t" squared so the position will be a parabala if you stretch its path horizontally)
Those two steps describe the way that the ball moves in the vertical direction, but since you threw the ball than the ball is moving in the x direction too, but since there is no gravity acting in this direction the equation to describe this motion is much simpler.
3- The equation he lists just says that you launch the ball from some initial position x(0), this will be 0m for 90% of problems since this is you are launchin your ball from what ever postion you are standing at, then you might as well say that you are standing at 0m position. Then when you launch the ball you give it some velocity in the horizontal direction. This equation allows you to determine the position horizontally.
This is a really simple problem in the large scheme of things, yet it describes everything from the way a ball moves, to how you move in the air after a jump, to rain falls. This stuff is enjoyable to a lot of people because its really amazing that you can rather acuately describe the movement of almost everything on earth with two relatively simple equations.