We first prove that $f(x) \geq 2 x$ for all $x>0$. Suppose, for the sake of contradiction, that $f(y)<2 y$ for some positive $y$. Choose $x$ such that $f((c+1) x+f(y))$ and $f(x+2 y)$ cancel out, that is,
$$(c+1) x+f(y)=x+2 y \Longleftrightarrow x=\frac{2 y-f(y)}{c}$$
Notice that $x>0$ because $2 y-f(y)>0$. Then $2 c x=0$, which is not possible. This contradiction yields $f(y) \geq 2 y$ for all $y>0$.

Now suppose, again for the sake of contradiction, that $f(y)>2 y$ for some $y>0$. Define the following sequence: $a_{0}$ is an arbitrary real greater than $2 y$, and $f\left(a_{n}\right)=f\left(a_{n-1}\right)+2 c x$, so that
$$\left\{ \begin{array}{r} (c+1) x+f(y)=a_{n} \\ x+2 y=a_{n-1} \end{array} \Longleftrightarrow x=a_{n-1}-2 y \quad \text{ and } \quad a_{n}=(c+1)\left(a_{n-1}-2 y\right)+f(y) .\right.$$
If $x=a_{n-1}-2 y>0$ then $a_{n}>f(y)>2 y$, so inductively all the substitutions make sense.

For the sake of simplicity, let $b_{n}=a_{n}-2 y$, so $b_{n}=(c+1) b_{n-1}+f(y)-2 y(*)$. Notice that $x=b_{n-1}$ in the former equation, so $f\left(a_{n}\right)=f\left(a_{n-1}\right)+2 c b_{n-1}$. Telescoping yields
$$f\left(a_{n}\right)=f\left(a_{0}\right)+2 c \sum_{i=0}^{n-1} b_{i} .$$
One can find $b_{n}$ from the recurrence equation $(*): b_{n}=\left(b_{0}+\frac{f(y)-2 y}{c}\right)(c+1)^{n}-\frac{f(y)-2 y}{c}$, and then
$$\begin{aligned} f\left(a_{n}\right) & =f\left(a_{0}\right)+2 c \sum_{i=0}^{n-1}\left(\left(b_{0}+\frac{f(y)-2 y}{c}\right)(c+1)^{i}-\frac{f(y)-2 y}{c}\right) \\ & =f\left(a_{0}\right)+2\left(b_{0}+\frac{f(y)-2 y}{c}\right)\left((c+1)^{n}-1\right)-2 n(f(y)-2 y) . \end{aligned}$$
Since $f\left(a_{n}\right) \geq 2 a_{n}=2 b_{n}+4 y$,
$$\begin{aligned} & f\left(a_{0}\right)+2\left(b_{0}+\frac{f(y)-2 y}{c}\right)\left((c+1)^{n}-1\right)-2 n(f(y)-2 y) \geq 2 b_{n}+4 y \\ = & 2\left(b_{0}+\frac{f(y)-2 y}{c}\right)(c+1)^{n}-2 \frac{f(y)-2 y}{c}, \end{aligned}$$
which implies
$$f\left(a_{0}\right)+2 \frac{f(y)-2 y}{c} \geq 2\left(b_{0}+\frac{f(y)-2 y}{c}\right)+2 n(f(y)-2 y),$$
which is not true for sufficiently large $n$.
A contradiction is reached, and thus $f(y)=2 y$ for all $y>0$. It is immediate that this function satisfies the functional equation.

After proving that $f(y) \geq 2 y$ for all $y>0$, one can define $g(x)=f(x)-2 x, g: \mathbb{R}_{>0} \rightarrow \mathbb{R}_{\geq 0}$, and our goal is proving that $g(x)=0$ for all $x>0$. The problem is now rewritten as
$$\begin{align*} & g((c+1) x+g(y)+2 y)+2((c+1) x+g(y)+2 y)=g(x+2 y)+2(x+2 y)+2 c x \\ \Longleftrightarrow & g((c+1) x+g(y)+2 y)+2 g(y)=g(x+2 y) \tag{1} \end{align*}$$
This readily implies that $g(x+2 y) \geq 2 g(y)$, which can be interpreted as $z>2 y \Longrightarrow g(z) \geq 2 g(y)$, by plugging $z=x+2 y$.

Now we prove by induction that $z>2 y \Longrightarrow g(z) \geq 2 m \cdot g(y)$ for any positive integer $2 m$. In fact, since $(c+1) x+g(y)+2 y>2 y, g((c+1) x+g(y)+2 y) \geq 2 m \cdot g(y)$, and by (??),
$$g(x+2 y) \geq 2 m \cdot g(y)+2 g(y)=2(m+1) g(y),$$
and we are done by plugging $z=x+2 y$ again.

The problem now is done: if $g(y)>0$ for some $y>0$, choose a fixed $z>2 y$ arbitrarily and and integer $m$ such that $m>\frac{g(z)}{2 g(y)}$. Then $g(z)<2 m \cdot g(y)$, contradiction.