The Generator point and Mod P in ECDSA I've been reading about The discrete logarithm problem as of recent and i decided to try it out on a small portion of numbers myself and i actually came to a  mental gridlock after watching this Video.
In the video, he uses the mod p which is prime to find a specific generator point (which is also prime) that could iterate all possible points(which i know that G is constant throughout the curve. but why use the order of the curve to find a suitable generator point.). And he also mentions that some numbers can't provide all possible digits that can iterate through all possible values.
But what baffles me the most is that the real values like the generator point and the order of the curves  for example secp256k1, secp192ri is/are not prime. And did i mention that all the answers on the discrete logarithm problem online ALL had prime generator points and the order of the curve were also prime. Can i get a breakdown as of what is going on?
Any opinions would be appreciated.
The video (at least, where the question links) illustrates the Baby-step/Giant-step Discrete Logarithm method in the multiplicative group modulo $p$, for prime $p$. That is the set $\{1,2,\ldots,p-2,p-1\}$, under the internal operation multiplication modulo $p$.
This group has essentially nothing to do with an Elliptic Curve group. The principle of Baby-step/Giant-step is the same regardless of group, however the operation analog to multiplication modulo $p$ is extremely different in an Elliptic Curve group.

It is important for security that the order of the group generated by $g$ (that is, the number of distinct values $g^x\bmod p$ can take) is prime, or at least has a large prime factor. When working in the multiplicative group modulo $p$, the order of the full group is $p-1$, and the order of $g$ is some divisor of that. One option is to choose $p$ such that $(p-1)/2$ is prime ($p$ is a so-called safe prime). Now we can take any $g$ except $g=1$ and $g=p-1$, and it's order will be $p-1$ or $(p-1)/2$, which both have a large prime factor $(p-1)/2$.
There is no imperious reason to choose a generator $g$ that is prime. It happens that $2$ and $3$ are popular, and prime; and some people have found satisfactory to choose $g$ prime.
There's even less reason to worry about prime $g$ in ECC groups. In secp256k1, both the $x$ and $y$ coordinates of the generator point are large even integers, thus not prime. What matters is the order of the group generated by $g$. That is noted $n$, and the curve parameters are usually chosen such that $n$ is prime.