What do these commutator identities have to do with the product rule for derivatives? 
This is a snapshot from the pdf I was reading. The last line here says that the identities above resemble the product rule for derivatives (which they do). But it is not explained in the pdf why.
Is the resemblance just a co-incidence, or is there some connection between the two concepts?
 A: The identities are expressing the product rule, not the chain rule, as already pointed out in the comments. To see that the identities are expressing the product rule, define the commutator operator $D_A: B\mapsto [A,B]$ acting on an operator $B$ and resulting in the operator $[A,B]$, being $A$ also a linear operator. Then the first identity can be restated as 
$$
D_\Omega(\Lambda\theta)=(D_\Omega\Lambda)\theta+\Lambda(D_\Omega\theta).
$$
The second identity does not add anything new, recalling that $D_AB=-D_BA$, the antisymmetric property of the commutator. Thus the operator $D_A$  is a derivation on the associative algebra of multiplication of linear operators. It is interesting to notice that $D_A$ is also a derivation on the commutator (Lie) algebra of linear operators, that is
$$
D_A[B,C]=[D_AB,C]+[B,D_AC],
$$
also known as Jacobi's identity.
A: It's not that commutators mimic differentiation; it's the reverse. Define $D:=\frac{d}{dx}$. The usual product rule $D(fg)=(Df)g+fDg$ can be rewritten as $[D,\,f]g=(Df)g$, or as the operator equation $[D,\,f]=Df$. Hence $$[D,\,fg]=D(fg)=(Df)g+fDg=[D,\,f]g+f[D,\,g].$$
Of course $[D,\,f]=Df$ looks odd at first, since it seems I'm saying $Df-fD=Df$ so that $fD=0$. No! The commutator $[D,\,f]$ is of the maps $g\mapsto Dg,\,g\mapsto f\cdot g$; the right-hand side $Df$ denotes $g\mapsto(Df)\cdot g.$ With both functions and $D$ thought of as operators on the vector space of functions, we see the usual product rule is a corollary of the commutator identity $$[A,\,BC]=[A,\,B]C+B[A,\,C]$$(which follows from $ABC-BCA=ABC-BAC+BAC-BCA$) together with the product rule $[D,\,f]=Df$.
