Why does RSA need p and q to be prime numbers? Despite having read What makes RSA secure by using prime numbers?, I seek clarification because I am still struggling to really grasp the underlying concepts of RSA.
Specifically,  why can't we choose a non-prime $p$ and $q$?
I do understand the key concept: multiplying two integers, even two very large integers is relatively simple. However, factoring a large integer is extremely difficult, even for a computer using known factoring algorithms.
If this is correct, then the original question posed by the OP in this thread continues to befuddle me.
This section from Henrick's answer in this post contains the relevant information:

It also means that if you were to select $p$, $q$ just as odd integers, you
would make it harder for yourself to find $\phi(n)$, while at the same time
decreasing the relative size of the second-largest prime factor, and
thereby making it easier for everyone else to factor n. In fact, it
would be as hard for you to factor n as it would be for everyone else,
so you would completely lose the trapdoor component of your scheme
(if not making it completely infeasible to find a pair $e$, $d$).

But I do not understand why we make it harder for ourselves to find $\phi(n)$. I BELIEVE that it has to do with the fact that for any prime, $n$, all integers up to $n-1$ are relatively prime. If the integer is not prime, then we actually need to find which integers up to n are actually relatively prime.
I understand how we are decreasing the relative size of the 2nd largest prime. For example: $10403$ has prime factors of $[101,103]$ while $11000$ has prime factors of $[2, 2, 2, 5, 5, 5, 11]$.
So, if I understand things correctly, choosing a non-prime $p$ and a non-prime $q$ would theoretically work, but the issue is that we would be creating a more insecure encryption scheme since:

the product of non-prime $p$ and non-prime $q$ are more easily factored (2nd largest prime is smaller than if $p$ and $q$ were prime);
finding $\phi(n)$ becomes more difficult for the key generator step, which decreases efficiency with the aforementioned decrease in security

Sorry if this is completely obvious, but I am a newbie in higher math and programming. I really want to understand this as deeply as I can.
However, factoring a large integer is extremely difficult, even for a computer using known factoring algorithms.


Not categorically. Factoring a large integer is trivial if it is only composed of small factors.

A fairly naive algorithm for factoring N is the following:

while N &gt; 1:
  for p in increasing_primes:
    while p divides N:
      N = N / p
      factors.add(p)


With this algorithm, $340$ $282$ $366$ $920$ $938$ $463$ $463$ $374$ $607$ $431$ $768$ $211$ $456$ can be factored in exactly $128$ iterations of the innermost while. (The number, of course, is $2^{128}$)

The more prime factors a composite number has, the smaller those factors have to be. For example, $919 \cdot 677 = 622$ $163$. With the naive algorithm, this takes $157 + 1 = 158$ iterations to factor. A number of roughly the same size comprised of three factors, $73 \cdot 89 \cdot 97 = 630$ $209$, only takes $25 + 2 = 27$ iterations to factor.

Similarly, a 1000-digit number composed of two roughly equally-sized factors will take about $10^{497}$ iterations to factor. A 1000-digit number composed of 100 roughly equally-sized factors will take about $434$ $294$ $481 + 99 \approx 10^{9}$ iterations. And a 1000-digit number composed of 1000 roughly equally-sized factors will take about $4 + 999 \approx 10^{3}$ iterations.

So, to make factoring as difficult as possible, you want the prime factors to be as large as possible, which means you want N to have the fewest factors possible. With $p$ and $q$ prime, $N$ has two factors. With $p$ and $q$ composite, $N$ has at least four factors, possibly more.

Note that this is far from a complete answer on why (we think) RSA is secure, but I hope it gives a more intuitive / example-based idea of why $p$ and $q$ should be primes.

Note also that $p$ and $q$ should be similar in magnitude. If one is much larger than the other, factorization becomes easier. For example, if $p = 43$, the naive algorithm would only take $14$ iterations.

(After reading the comments (thanks Taemyr and Daz C), I corrected some mistakes in the answer. The main point is unaffected.)