Why the resultant period of two signals is the LCM of the individual period of that signal?

Take two sine waves of periods \$T_1 = 2s\$ and \$T_2 = 3s\$.

Suppose they both start at time = 0s.

Then their "end of the cycle" points coincide only at the multiples of \$LCM(T_1,T_2) = LCM(2,3) = 6s \$

While adding both signals, we take summation of all respective points of both signals at a time instant, and we get a set of non-repeating values between 0 to 6s interval. After that, these values roll over and repeat every 6s. Hence the period of the resulting signal becomes 6s.

This is just basic math.

It may be easier to think of this by considering the period rather than the frequency. If you have a signal with 1 second period and add to it a signal with ½ second period, for example, with what period will the combined signal repeat. The answer is obviously 1 second. Another way to look at this is that the ½ second signal is a harmonic of the 1 second signal. Adding harmonics doesn't change the fundamental frequency.

For the harmonic addition logic to be valid, added signals have to actually be harmonics. That means their frequencies must be integer multiples of the fundamental. Therefore, the problem is finding the fundamental of a set of frequencies. The fundamental is the highest frequency that all the others are integer multiples of. What you are looking for is therefore the largest common denominator. Note that this largest common denominator of the frequency results in the same answer as the least common multiple of the period, which is what you seem to be asking about.

Did you watch the entire video? The first part, leading up to the frame you show, perfectly explains the concept of the LCM period. The frame you show is simply a few special cases in which the LCM period happens to be equal to the period of the slowest signal.

Later, it turns out that this kind of special case is important for developing Fourier analysis.