If $f,g: \mathbb{R} \to \mathbb{R}$ are functions that satisfy $f(x+g(y)) = 2x+y $ $\forall x,y \in \mathbb{R}$, then determine $g(x+f(y))$.
1. Given the functional equation \( f(x + g(y)) = 2x + y \) for all \( x, y \in \mathbb{R} \), we need to determine \( g(x + f(y)) \).

2. First, we will find an expression for \( f \). Let's plug in \( y = 0 \) into the given equation:
   \[
   f(x + g(0)) = 2x + 0 \implies f(x + g(0)) = 2x.
   \]
   Let \( x = x - g(0) \). Then,
   \[
   f((x - g(0)) + g(0)) = 2(x - g(0)) \implies f(x) = 2x - 2g(0).
   \]

3. Now, we have \( f(x) = 2x - 2g(0) \). Next, we substitute this expression back into the original functional equation:
   \[
   f(x + g(y)) = 2x + y.
   \]
   Using \( f(x) = 2x - 2g(0) \), we get:
   \[
   2(x + g(y)) - 2g(0) = 2x + y \implies 2x + 2g(y) - 2g(0) = 2x + y.
   \]
   Simplifying, we find:
   \[
   2g(y) - 2g(0) = y \implies 2g(y) = y + 2g(0) \implies g(y) = \frac{y}{2} + g(0).
   \]

4. Now, we need to determine \( g(x + f(y)) \). Using the expressions for \( f \) and \( g \), we have:
   \[
   g(x + f(y)) = g\left(x + (2y - 2g(0))\right).
   \]
   Substituting \( g \) into this, we get:
   \[
   g(x + 2y - 2g(0)) = \frac{x + 2y - 2g(0)}{2} + g(0).
   \]
   Simplifying, we obtain:
   \[
   g(x + f(y)) = \frac{x}{2} + y - g(0) + g(0) = \frac{x}{2} + y.
   \]

The final answer is \(\boxed{\frac{x}{2} + y}\).