Suppose we have a function defined like so;
$$f_n( x, x ) = f_{n+1}( x, 2 )$$
$$f_n( x, x, x ) = f_{n+1}( x, 3 )$$
$$f_n( x, x, x, x,\dots) = f_{n+1}( x, \text{number of} x\text{'s} )$$
$f_1( x, y ) = x + y$
We know that $f_2(f_1( x, y ), a ) = f_1( f_2( x, a ), f_2( y, a ) )$ because of the distributive property, but do all functions have this property?
Does $$f_{n+1}( f_n( x, y ),a ) = f_n( f_{n+1}( x, a ), f_{n+1}( y, a ) )$$ for any $n$?
I'll use infix notation for the hyperoperations:
$$a\times_0b=1+b$$ $$a\times_1b=a+b$$ $$a\times_2b=ab$$ $$a\times_3b=a^b$$ $$\qquad\qquad\qquad\qquad\quad a\times_4b=a^{(a\times_4(b-1))},\quad a\times_41=a$$
We know that multiplication distributes over addition, on both sides. Exponentiation distributes over multiplication on the right but not on the left:
$$(x\cdot y)^a=(x^a)\cdot(y^a)$$ $$\qquad\qquad a^{(x\cdot y)}\neq(a^x)\cdot(a^y)=a^{(x+y)}$$
And addition distributes over succession:
$$(x\times_0y)\times_1a=(1+y)+a$$ $$(x\times_1a)\times_0(y\times_1a)=1+(y+a)$$ $$a\times_1(x\times_0y)=a+(1+y)$$ $$(a\times_1x)\times_0(a\times_1y)=1+(a+y)$$
What about tetration?
$$(x^y)\times_4a\overset?=(x\times_4a)^{(y\times_4a)}$$ $$a\times_4(x^y)\overset?=(a\times_4x)^{(a\times_4y)}$$
It's not right-distributive, as we can see with $a=2,x=2,y=3$:
$$(2^3)\times_42\overset?=(2\times_42)^{(3\times_42)}$$ $$(2^3)^{(2^3)}\overset?=(2^2)^{(3^3)}$$ $$2^{(3\cdot2^3)}\overset?=2^{(2\cdot3^3)}$$ $$2^{24}\neq2^{54}$$
And it's not left-distributive either, as we can see with $a=2,x=1,y=1$:
$$2\times_4(1^1)\overset?=(2\times_41)^{(2\times_41)}$$ $$2\neq2^2=4$$