As you ask for a hint, here are two hints to help you along. Below is a sketch of a full proof. Let me know when I can 'unhide' all the hidden text to make the answer more legible for future readers.
Hint 1:
For all $a,b\in\Bbb{R}$ find $c\in\Bbb{R}$ such that the second identity becomes of the form
$$P(u,-u)+P(v,-v)+P(w,-w)=0.$$
Hint 2:
Deduce that if $\deg{P}>1$ then $P$ is divisible by $X+Y$.
Full solution: The polynomials that satisfy the conditions are precisely the polynomials
$$(X-2Y)(X+Y)^n,$$
with $n\in\Bbb{N}$. It is not hard to verify that these polynomials satisfy the conditions. Showing that there are no other solutions is more work. Below is a proof is by induction on the degree.
Observation 1: The unique solution $P\in\Bbb{R}[X,Y]$ with $\deg P\leq1$ is $P=X-2Y$.
Proof.
There are no constant solutions, and for $n=1$ setting $P=uX+vY$ shows that
$$(2u+v)(a+b+c)=0,$$
holds for all $a,b,c\in\Bbb{R}$,
and together with $P(1,0)=1$ this implies $P=X-2Y$.$\hspace{10pt}\square$
Observation 2: If $P\in\Bbb{R}[X,Y]$ satisfies the conditions and $\deg P>1$ then $X+Y$ divides $P$.
Proof. Suppose $P\in\Bbb{R}[X,Y]$ satisfies the conditions and $\deg P>1$. Plugging in $c=-a-b$ shows that for all $a,b\in\Bbb{R}$ we have
$$0=P(a+b,-a-b)+P(-a,a)+P(-b,b)=((a+b)^n+(-a)^n+(-b)^n)P(1,-1),$$
which implies that $P(1,-1)=0$ because $n>1$, and hence that $P(X,-X)=0$.
This means $P$ is divisible $X+Y$.$\hspace{10pt}\square$
Proof of full solution. Now we can prove by induction that for all $n\in\Bbb{N}$ we have
If $P\in\Bbb{R}[X,Y]$ satisfies the conditions and $\deg P=n+1$ then $P=(X-2Y)(X+Y)^n$.
The base case $n=0$ is covered by observation 1. So let $n\in\Bbb{N}$ and suppose that the statement above holds for $n$.
Suppose $P\in\Bbb{R}[X,Y]$ satisfies the conditions and $\deg P=n+2$. Then $P$ is divisible by $X+Y$ by observation 2, which means there exists $Q\in\Bbb{R}[X,Y]$ such that $P=(X+Y)Q$. Then clearly $\deg Q=n+1$, and we verify that $Q$ also satisfies the conditions:
- Because $P$ and $X+Y$ are homogeneous, also $Q$ is homogeneous.
- For all $a,b,c\in\Bbb{R}$ we have
\begin{eqnarray*}
0&=&P(a+b,c)+P(b+c,a)+P(c+a,b)\\
&=&(a+b+c)(Q(a+b,c)+Q(b+c,a)+Q(c+a,b)),
\end{eqnarray*}
which shows that for all $a,b,c\in\Bbb{R}$ with $a+b+c\neq0$ we have
$$Q(a+b,c)+Q(b+c,a)+Q(c+a,b)=0.$$
Because $Q$ is a polynomial, it follows that this holds for all $a,b,c\in\Bbb{R}$.
- Clearly $P(1,0)=1$ implies $Q(1,0)=1$.
This shows that $Q$ satisfies the conditions and $\deg Q=n+1$, so by induction hypothesis
$$Q=(X-2Y)(X+Y)^n
\qquad\text{ and hence }\qquad
P=(X-2Y)(X+Y)^{n+1},$$
which completes the proof by induction.
Note that
$$g(x) + h(y) = f(x + y) = f(y + x) = g(y) + h(x) \implies h(x) - g(x) = h(y) - g(y)$$
for all $x, y \in \Bbb{R}$. That is, $h - g$ is a constant function, i.e. there exists some $k \in \Bbb{R}$ such hat $h(x) = g(x) + k$ for all $x \in \Bbb{R}$.
This gives us the equivalent functional equation
$$f(x + y) = g(x) + g(y) + k.$$
Note that, when $y = 0$, we simply see that
$$f(x) = g(x) + g(0) + k,$$
hence
$$g(x + y) + g(0) + k = g(x) + g(y) + k \implies g(x + y) + g(0) = g(x) + g(y).$$
Let $L(x) = g(x) - g(0)$. Then, the above equation simplifies to
$$L(x + y) = L(x) + L(y),$$
which is Cauchy's functional equation. Since $g$ is continuous, so is $L$, and hence $L$ is linear. On $\Bbb{R}$, this means $L(x) = ax$ for some $a \in \Bbb{R}$.
So, rebuilding, we have
\begin{align*}
g(x) &= ax + c \\
h(x) &= ax + c + k \\
f(x) &= ax + 2c + k,
\end{align*}
where $a, c, k \in \Bbb{R}$ are parameters. Checking this family of possible solutions, we get
$$f(x + y) = a(x + y) + 2c + k = ax + c + ay + c + k = g(x) + h(y),$$
verifying that all functions of the above form are indeed solutions, yielding a complete family of solutions.
Best Answer
The sequence goes like this:
$$ \{ n, f(n) , f \circ f(n)... \}$$
The action of 'composition' with $f$ generates the next element in the sequence. Note that we are not making any assumptions about the nature of $f$ when we say that the first element of the said sequence is $n$, so everything is still done in total generality.