Determine all functions $ f: \mathbb{N}_0 \rightarrow \mathbb{N}_0$ such that $ f(f(n))\plus{}f(n)\equal{}2n\plus{}6$ for all $ n \in \mathbb{N}_0$.
1. Define a sequence \( a_0 = R \) for some \( R \in \mathbb{N}_0 \), and \( a_{m+1} = f(a_m) \) for all \( m \in \mathbb{N}_0 \).

2. From the given functional equation \( f(f(n)) + f(n) = 2n + 6 \), we can write the relation for the sequence:
   \[
   a_{m+2} + a_{m+1} = 2a_m + 6
   \]

3. Also, we have:
   \[
   a_{m+1} + a_m = 2a_{m-1} + 6
   \]

4. Let \( d_m = a_m - a_{m-1} \). Taking the difference between the above recursions yields:
   \[
   d_{m+2} + d_{m+1} = 2d_m
   \]

5. The characteristic polynomial of this recursion is:
   \[
   \lambda^2 + \lambda - 2 = 0
   \]
   Solving this quadratic equation, we get the roots:
   \[
   \lambda = -2 \quad \text{and} \quad \lambda = 1
   \]

6. Thus, the general solution for \( d_m \) is:
   \[
   d_m = P(-2)^m + Q(1)^m = P(-2)^m + Q
   \]
   for some constants \( P, Q \in \mathbb{R} \).

7. Therefore, the sequence \( a_m \) can be written as:
   \[
   a_m = \sum_{k=1}^m d_k + a_0 = \sum_{k=1}^m (P(-2)^k + Q) + a_0
   \]

8. Simplifying the sum, we get:
   \[
   a_m = P \sum_{k=1}^m (-2)^k + Qm + a_0
   \]
   The sum of the geometric series \( \sum_{k=1}^m (-2)^k \) is:
   \[
   \sum_{k=1}^m (-2)^k = -2 \frac{(-2)^m - 1}{-2 - 1} = \frac{2^{m+1} - 2}{3}
   \]
   Therefore:
   \[
   a_m = P \left( \frac{2^{m+1} - 2}{3} \right) + Qm + a_0
   \]

9. If \( P \neq 0 \), then there exists some \( M \in \mathbb{N}_0 \) such that \( a_M < 0 \). This is a contradiction since \( f: \mathbb{N}_0 \rightarrow \mathbb{N}_0 \).

10. Thus, \( P = 0 \), and:
    \[
    a_m = Qm + a_0
    \]

11. So, \( a_{m+1} = Q(m+1) + a_0 \) and \( a_{m+2} = Q(m+2) + a_0 \). Substituting these into the original relation:
    \[
    Q(m+2) + a_0 + Q(m+1) + a_0 = 2(Qm + a_0) + 6
    \]
    Simplifying, we get:
    \[
    Qm + 2Q + a_0 + Qm + Q + a_0 = 2Qm + 2a_0 + 6
    \]
    \[
    2Qm + 3Q + 2a_0 = 2Qm + 2a_0 + 6
    \]
    \[
    3Q = 6 \implies Q = 2
    \]

12. Therefore, \( a_m = 2m + a_0 \). Since \( a_0 = R \), we have:
    \[
    a_m = 2m + R
    \]

13. For \( f(n) \), we have \( f(n) = 2n + R \). Plugging this into the original equation:
    \[
    f(f(n)) = f(2n + R) = 2(2n + R) + R = 4n + 3R
    \]
    \[
    f(f(n)) + f(n) = 4n + 3R + 2n + R = 6n + 4R
    \]
    \[
    2n + 6 = 6n + 4R
    \]
    \[
    4R = 6 - 6n
    \]
    Since this must hold for all \( n \), we conclude \( R = 2 \).

14. Therefore, the function is:
    \[
    f(n) = n + 2
    \]

The final answer is \( \boxed{ f(n) = n + 2 } \).