As Marktmeister has mentioned in a comment, for the functional equation
$$ f \bigl ( f ( x ) - x \bigr ) = 2 x \tag 0 \label 0 $$
to make sense for a given $ f : \mathbb R ^ + _ 0 \to \mathbb R ^ + _ 0 $ and all $ x \in \mathbb R ^ + _ 0 $, one must have $ f ( x ) \ge x $ for all $ x \in \mathbb R ^ + _ 0 $. In particular, $ f \bigl ( f ( x ) - x \bigr ) \ge f ( x ) - x $ for all $ x \in \mathbb R ^ + _ 0 $, which by \eqref{0} gives $ f ( x ) \le 3 x $ for all $ x \in \mathbb R ^ + _ 0 $. Let's iterate this method: define the sequences $ ( a _ n ) _ { n = 0 } ^ \infty $ and $ ( b _ n ) _ { n = 0 } ^ \infty $ of positive real numbers recursively by $ a _ 0 = 1 $ and for any nonnegative integer $ n $, $ b _ n = \frac 2 { a _ n } + 1 $ and $ a _ { n + 1 } = \frac 2 { b _ n } + 1 $. Note that similar to what we did above, by \eqref{0} and mathematical induction, one gets $ a _ n x \le f ( x ) \le b _ n x $ for all $ x \in \mathbb R ^ + _ 0 $ and all positive integers $ n $. Therefore, if we prove that as $ n $ tends to infinity, $ a _ n $ increasingly goes to $ 2 $ and $ b _ n $ decreasingly goes to $ 2 $, then we end up with $ f ( x ) = 2 x $ for all $ x \in \mathbb R ^ + _ 0 $, which indeed gives a solution to \eqref{0}. For this purpose, first use mathematical induction to prove that for any nonnegative integer $ n $, $ 1 \le a _ n < 2 $ and $ 2 < b _ n \le 3 $. Then, note that for any nonnegative integer $ n $,
$$ 0 < 2 - a _ { n + 1 } = 2 - \left ( \frac 2 { \frac 2 { a _ n } + 1 } + 1 \right ) = \frac { 2 - a _ n } { 2 + a _ n } \le \frac { 2 - a _ n } { 2 + 1 } = \frac { 2 - a _ n } 3 $$
and
$$ 0 < b _ { n + 1 } - 2 = \left ( \frac 2 { \frac 2 { b _ n } + 1 } + 1 \right ) - 2 = \frac { b _ n - 2 } { b _ n + 2 } < \frac { b _ n - 2 } { 2 + 2 } = \frac { b _ n - 2 } 4 \text . $$
Hence, by mathematical induction, we get
$$ 0 < 2 - a _ n \le \frac { 2 - a _ 0 } { 3 ^ n } = \frac 1 { 3 ^ n } $$
and
$$ 0 < b _ n - 2 \le \frac { b _ 0 - 2 } { 4 ^ n } = \frac 1 { 4 ^ n } $$
for all nonnegative integers $ n $, which proves what was desired.
Concerning the part you added later, if $ f : \mathbb R ^ + _ 0 \to \mathbb R ^ + _ 0 $ satisfies
$$ f \bigl ( f ( x ) - \alpha x \bigr ) = \beta x \tag 1 \label 1 $$
for some constants $ \alpha , \beta \in \mathbb R ^ + $ and all $ x \in \mathbb R ^ + _ 0 $, the same method as above works. Note that for all $ x \in \mathbb R ^ + _ 0 $ and all $ \gamma \in \mathbb R ^ + $, $ f \bigl ( f ( x ) - \alpha x \bigr ) \ge \gamma \bigl ( f ( x ) - \alpha x \bigr ) $ implies $ f ( x ) \le \left ( \frac \beta \gamma + \alpha \right ) x $ (by \eqref{1}), and $ f \bigl ( f ( x ) - \alpha x \bigr ) \le \gamma \bigl ( f ( x ) - \alpha x \bigr ) $ implies $ f ( x ) \ge \left ( \frac \beta \gamma + \alpha \right ) x $. Therefore, similar to before, we can define $ ( a _ m ) _ { m = 0 } ^ \infty $ and $ ( b _ m ) _ { m = 0 } ^ \infty $ by $ a _ 0 = \alpha $, $ b _ m = \frac \beta { a _ m } + \alpha $ and $ a _ { m + 1 } = \frac \beta { b _ m } + \alpha $, and inductively show that $ a _ m x \le f ( x ) \le b _ m x $ for all $ x \in \mathbb R ^ + $ and all nonnegative integers $ m $. Taking limits, you can then find out that $ f ( x ) = c x $ gives the unique solution, where $ c $ is the positive root of $ c ^ 2 - \alpha c - \beta $, i.e. $ c = \frac { \alpha + \sqrt { \alpha ^ 2 + 4 \beta } } 2 $. I leave verifying the details to you. In the special case where $ \alpha = n $ and $ \beta = n + 1 $, we have $ c = n + 1 $.
Best Answer
For the first one : If $x=y=0$, then $f(0)=0$, if $x=y$, then $f(0)=2xf(x)$, so $\forall x, 2xf(x)=0$, so $\forall x, f(x)=0$. So $f=0$.
For the second one : $f(f(f(x)))=f(x+1)$ and $f(f(f(x)))=f(x)+1$. So $f(x+1)=f(x)+1$. So $\forall x,f(x)=f(0)+x$. Then note that $f(f(0))=0+1=1$ with the property in the question and $f(f(0))=f(f(0))+1$ with the previous equality, so $1=f(f(0))=f(f(0))+1=1+1=2$. It is absurd, so there is no such $f$.