What you're arguing is that if you pick a uniformly random element of a non-prime order group and you multiply by the cofactor, you'll get a uniformly random element of the prime-order subgroup. This is true and you could prove it by arguing that cosets of the kernel of this multiplication form a partition of the group.
However, if a cyclic group $C$ has order $np$ for $p$ prime and $gcd(n,p)=1$, then it can be written as $C\cong C_n\times C_p$, i.e. a product of two cyclic groups. So any element $g\in C$ can be written as a pair $(g_n,g_p)$. Multiplying just by the cofactor $n$ maps us from $(g_n,g_p)$ to $(0,ng_p)$. This cleared the cofactor part, but it changed the prime-order part. What if we didn't want that? What if we wanted to go from $(g_n,g_p)$ to $(0,g_p)$? We want to project to one component (almost like multiplying by $(0,1)$, but these are additive groups so we can't multiply elements like that). One way to see this is that after we multiply by $n$ (which clears out the first component) we need to multiply by something to clear out $n$ in the second component. Since $gcd(n,p)=1$, there is some $n'\equiv n^{-1}\mod p$, so multiplying by $nn'$ does the trick.
We can see this another way by noting that we have a projection $P:C_n\times C_p\rightarrow C_n\times C_p$ such that $P(a,b)=(0,b)$, and we want to create a map $\tilde{P}:C\rightarrow C$ such that $\tilde{P}$ has the same action as $P$ given the isomorphism from $C_n\times C_p$ to $C$. The original posted gives an explicit isomorphism in the comments, as $f(a,b)=an + bp$. You can then show that multiplying by $nn'$ in $C$ does have the effect of projecting to $C_p$ in $C_n\times C_p$.
But I have no idea why we would want that. I tried reading through the rest of it and I still don't really see why. Here's my best guess: later on they say for various reasons we want to output public keys as uniformly random points on Curve25519, even though we only do the key exchange on the prime-order subgroup. So, we need to add a random point in the cofactor group. That is, Alice's "real" public key would be some point $(0,p_A)=(0,s_Ap)$, but she modifies it to $(r_A,p_A)$ by adding a random cofactor point $r_A$ (here using the product notation of above).
The way they explain it, both parties have already multiplied a cofactor of 4 into their private keys, so multiplying by their private key will automatically clear out the cofactor part of the group, e.g., if Bob's secret key $s_B$ is divisible by $4$, then $s_B(r_A,p_A)=(s_Br_A,s_Bp_A)=(0,s_Bp_A)$, as it should be.
Suppose instead I stored my secret key as $s_B'\in \{0,1,\dots,p-1\}$, so that my public key is $(0,s_B'p)$. Then if I multiplied by this secret key, I'd get $(s_B'r_A,s_B'p_A)$ where $s_B'r_A$ is not necessarily $0$. I want to clear that out without modifying $s_B'p_A$ in any way. So, I use the more complicated cofactor clearing shown in the post.
It seems to me like a simpler way to do the same would be to enforce multiplying by the cofactor at the start of any scalar multiplication, but there might be subtle issues I'm missing, or I'm totally wrong about why we would do this.
