Assume that $f : \mathbb{R}^+ \to \mathbb{R}^+$ and the polynomial $g$ with non-negative coefficients and $g(0) = 0$ satisfy the conditions of the problem. For positive reals with $x > y$, we shall write $P(x, y)$ for the relation:
$$f(f(x) + g(y)) = f(x - y) + 2y.$$
1. Step 1. $f(x) \ge x$. Assume that this is not true. Since $g(0) = 0$, $g(x) + x$ is injective on positive reals. If $f(x) < x$ for some positive real $x$, then setting $y$ such that $y + g(y) = x - f(x)$ (where obviously $y < x$), we shall get $f(x) + g(y) = x - y$ and by $P(x, y)$, $f(f(x) + g(y)) = f(x - y) + 2y$, we get $2y = 0$, a contradiction.

2. Step 2. $g(x) = cx$ for some non-negative real $c$. We will show $\deg g \le 1$ and together with $g(0) = 0$ the result will follow. Assume the contrary. Hence there exists a positive $l$ such that $g(x) \ge 2x$ for all $x \ge l$. By Step 1 we get
$$\forall x > y \ge l : f(x - y) + 2y = f(f(x) + g(y)) \ge f(x) + g(y) \ge f(x) + 2y$$
and therefore $f(x - y) \ge f(x)$. We get $f(y) \ge f(2y) \ge \dots \ge f(ny) \ge ny$ for all positive integers $n$, which is a contradiction.

3. Step 3. If $c \ne 0$, then $f(f(x) + y + c^2 + 2) = f(x + 1) + y + 2c$. Indeed by $P(f(x + \frac{y}{2} + 1) + \frac{cy}{2} + c, c)$, we get
$$f(f(f(x + \frac{y}{2} + 1) + \frac{cy}{2} + c) + c^2) = f(f(x + \frac{y}{2} + 1) + \frac{cy}{2}) + 2c = f(x + 1) + y + 2c.$$
On the other hand by $P(x + \frac{y}{2} + 1, \frac{y}{2} + 1)$, we have:
$$f(x) + y + 2 = f\left(f\left(x + \frac{y}{2} + 1\right) + g\left(\frac{y}{2} + 1\right)\right) = f\left(f\left(x + \frac{y}{2} + 1\right) + \frac{cy}{2} + c\right).$$
Substituting in the LHS of $P(f(x + \frac{y}{2} + 1) + \frac{cy}{2} + c, c)$, we get $f(f(x) + y + 2 + c^2) = f(x + 1) + y + 2c$.

4. Step 4. There is $x_0$, such that $f(x)$ is linear on $(x_0, \infty)$. If $c \neq 0$, then by Step 3, fixing $x=1$, we get $f(y + f(1) + 2 + c^2) = y + f(2) + 2c$ which implies that $f$ is linear for $y > f(1) + 2 + c^2$. As for the case $c = 0$, consider $y, z \in (0, \infty)$. Pick $x > \max(y, z)$, then by $P(x, x - y)$ and $P(x, x - z)$ we get:
$$f(y) + 2(x - y) = f(f(x)) = f(z) + 2(x - z)$$
which proves that $f(y) - 2y = f(z) - 2z$ and therefore $f$ is linear on $(0, \infty)$.

5. Step 5. $g(y) = y$ and $f(x) = x$ on $(x_0, \infty)$. By Step 4, let $f(x) = ax + b$ on $(x_0, \infty)$. Since $f$ takes only positive values, $a \ge 0$. If $a = 0$, then by $P(x + y, y)$ for $y > x_0$ we get:
$$2y + f(x) = f(f(x + y) + g(y)) = f(b + cy).$$
Since the LHS is not constant, we conclude $c \neq 0$, but then for $y > x_0/c$, we get that the RHS equals $b$ which is a contradiction.
Hence $a > 0$. Now for $x > x_0$ and $x > (x_0 - b)/a$ large enough by $P(x + y, y)$ we get:
$$ax+b+2y = f(x)+2y = f(f(x+y)+g(y)) = f(ax+ay+b+cy) = a(ax+ay+b+cy)+b.$$
Comparing the coefficients before $x$, we see $a^2 = a$ and since $a \neq 0$, $a = 1$. Now $2b = b$ and thus $b = 0$. Finally, equalising the coefficients before $y$, we conclude $2 = 1 + c$ and therefore $c = 1$.
Now we know that $f(x) = x$ on $(x_0, \infty)$ and $g(y) = y$. Let $y > x_0$. Then by $P(x + y, x)$ we conclude:
$$f(x) + 2y = f(f(x + y) + g(y)) = f(x + y + y) = x + 2y.$$
Therefore $f(x) = x$ for every $x$. Conversely, it is straightforward that $f(x) = x$ and $g(y) = y$ do indeed satisfy the conditions of the problem. □

Assume that the function $f: \mathbb{R}^+ \to \mathbb{R}^+$ and the polynomial with non-negative coefficients $g(y) = yg_1(y)$ satisfy the given equation. Fix $x = x_0 > 0$ and note that:
$$f(f(x_0 + y) + g(y)) = f(x_0 + y - y) + 2y = f(x_0) + 2y.$$
Assume that $g = 0$. Then $f(f(x + y)) = f(x) + 2y$ for $x, y > 0$. Let $x > 0$ and $z > 0$. Pick $y > 0$. Then:
$$2y + f(x + z) = f(f(x + y + z)) = f(f(x + z + y)) = f(x) + 2(z + y).$$
Therefore $f(x+z) = f(x) + 2z$ for any $x > 0$ and $z > 0$. Setting $c = f(1)$, we see that $f(z+1) = c + 2z$ for all positive $z$. Therefore if $x, y > 1$ we have that $f(x+y) = c + 2(x+y-1) > 1$. This shows that:
$$f(f(x+y)) = c + 2(f(x+y) - 1) = 3c + 4(x+y) - 4.$$
On the other hand $f(x)+2y = c+2x+2y$. Therefore the equality $f(f(x+y)) = f(x)+2y$ is not universally satisfied.

From now on, we assume that $g \neq 0$. Therefore $g$ is strictly increasing with $g(0) = 0$, $\lim_{y \to \infty} g(y) = \infty$, i.e. $g$ is bijective on $[0, \infty)$ and $g(0) = 0$.
Let $x > 0, y > 0$ and set $u = f(x+y), v = g(y)$. From above, we have $u > 0$ and $v > 0$. Therefore:
$$f(f(u+v) + g(v)) = f(u) + 2v = f(f(x+y)) + 2g(y).$$
On the other hand $f(u+v) = f(f(x+y) + g(y)) = f(x) + 2y$. Therefore we obtain that:
$$f(f(x) + 2y + g(g(y))) = f(f(x+y)) + 2g(y).$$
Since $g$ is bijective from $(0, \infty)$ to $(0, \infty)$ for any $z > 0$ there is $t$ such that $g(t) = z$. Applying this observation to $z = g(g(y)) + 2y$ and setting $x' = x + t$, we obtain that:
$$f(f(x+t+y))+2g(y) = f(f(x'+y))+2g(y) = f(f(x')+g(g(y))+2y) = f(f(x+t)+g(t)) = f(x)+2t.$$
Thus if we denote $h(y) = g(g(y)) + 2y$, then $t = g^{-1}(h(y))$ and the above equality can be rewritten as:
$$f(f(x+g^{-1}(h(y))+y)) = f(x)+2g^{-1}(h(y))-2g(y) = f(x)+2g^{-1}(h(y))+2y-2y-2g(y).$$
Let $s(y) = g^{-1}(h(y)) + y$ and note that since $h$ is continuous and monotone increasing, $g$ is continuous and monotone increasing, then so are $g^{-1}$ and consequently $g^{-1} \circ h$ and $s$. It is also clear, that $\lim_{y \to 0} s(y) = 0$ and $\lim_{y \to \infty} s(y) = \infty$. Therefore $s$ is continuously bijective from $[0, \infty)$ to $[0, \infty)$ with $s(0) = 0$.
Thus we have:
$$f(f(x + s(y))) = f(x) + 2s(y) - 2y - 2g(y)$$
and using that $s$ is invertible, we obtain:
$$f(f(x + y)) = f(x) + 2y - 2s^{-1}(y) - 2g(s^{-1}(y)).$$
Setting $y = x_0$, we get:
$$f(x) + 2x_0 - 2s^{-1}(x_0) - 2g(s^{-1}(x_0)) = f(x_0) + 2x - 2s^{-1}(x) - 2g(s^{-1}(x)).$$
Since this equality is valid for any $x > x_0$ we actually have that:
$$f(x) - 2x + 2s^{-1}(x) + 2g(s^{-1}(x)) = c \text{ for some fixed constant } c \in \mathbb{R} \text{ and all } x \in \mathbb{R}^+.$$
Let $\phi(x) = -x + 2s^{-1}(x) + 2g(s^{-1}(x))$. Then:
$$f(f(x+y)+g(y)) = f(x+y+\phi(x+y)+c+g(y)) = x+y+g(y)+\phi(x+y)+2c+\phi(x+y+\phi(x+y)+c+g(y)).$$
On the other hand:
$$f(f(x + y) + g(y)) = f(x) + 2y = x + \phi(x) + 2y + c.$$
Therefore:
$$g(y) + \phi(x + y) + c + \phi(x + y + \phi(x + y) + c + g(y)) = \phi(x) + y + c.$$
Noting that $\phi$ is continuous on $[0, \infty)$, since it is sum of continuous functions, and letting $y$ tend to 0, we obtain that:
$$\phi(x) + c + \phi(x + \phi(x) + c) = \phi(x).$$
Therefore $\phi(x + \phi(x) + c) + c = 0$ and substituting in the definition of $\phi(x) = -x + 2s^{-1}(x) + 2g(s^{-1}(x))$ we obtain:
$$-x - \phi(x) - c + c + 2s^{-1}(x + \phi(x) + c) + 2g(s^{-1}(x + \phi(x) + c)) = 0.$$
Consequently:
$$c + 2s^{-1}(x + \phi(x) + c) + 2g(s^{-1}(x + \phi(x) + c)) = x + \phi(x) + c.$$
Thus:
$$\phi(c + 2s^{-1}(x + \phi(x) + c) + 2g(s^{-1}(x + \phi(x) + c))) = \phi(x + \phi(x) + c) = -c.$$
Finally note that $x + \phi(x) + c = 2s^{-1}(x) + 2g(s^{-1}(x)) + c =: u(x)$ and since $g$ and $s^{-1}$ are monotone and bijective on $[0, \infty)$, $u(x)$ exhausts $[c, \infty)$ when $x$ ranges on $[0, \infty)$. It follows that $\phi(x) = -c$ for $x \in [c, \infty)$. It follows that for $x > \max(c, 0)$:
$$f(x) = x + c - \phi(x) = x - 2c.$$
In particular, since $f(x) > 0$, $c \le 0$. Now for $x > \max(c, 0)$ and $y > 0$ we have:
$$x + 2y - 2c = f(x) + 2y = f(f(x + y) + g(y)) = f(x + y - 2c + g(y)) = x + y + g(y) - 4c.$$
Since this is valid for any $y$, we conclude $g(y) = y$ and $c = 0$. Now it follows that $f(x) = x$ for $x \in (0, \infty)$.
It is also straightforward to check that $f(x) = x$ and $g(y) = y$ satisfy the equality:
$$f(f(x + y) + g(y)) = f(x + 2y) = x + 2y = f(x) + 2y.$$