All solutions satisfying the problem are $f(x) = x^2 + c, g(x) = x$, where $c$ is an arbitrary real number.

Observe that the left side of the original equation is symmetric in $x, y$, so substituting $(y, x)$ gives
$$g(f(x + y)) = f(y) + (2y + x)g(x)$$
Subtracting this from the original equation yields
$$f(x) + (2x + y)g(y) = f(y) + (2y + x)g(x)$$
Substituting $(x, 0), (1, x)$ into this equation gives
$$f(x) + 2xg(0) = f(0) + xg(x)$$
$$f(1) + (2 + x)g(x) = f(x) + (2x + 1)g(1)$$
Adding these two equations and rearranging gives
$$g(x) = x(g(1) - g(0)) + \frac{g(1) + f(0) - f(1)}{2}$$
Since $g(0), g(1), f(0), f(1)$ are all constants, $g(x)$ must be a constant or a first-degree polynomial. From
$$f(x) + 2xg(0) = f(0) + xg(x)$$
we can further see that $f(x)$ must be a polynomial of degree at most two.

Let $g(x) = Ax + B$ and substitute back into the original problem to get
$$Af(x + y) + B = f(x) + (2x + y)(Ay + B)$$

Substituting $(x, 0)$ gives
$$(A - 1)f(x) = (2x - 1)B$$
If $A \neq 1$, then
$$f(x) = \frac{(2x - 1)B}{A - 1}$$
Substituting this back into the original problem shows it does not work, a contradiction! Therefore $A = 1$, and hence $B = 0$, that is $g(x) = x$.

Substituting $g(x) = x$ into
$$f(x) + 2xg(0) = f(0) + xg(x)$$
gives
$$f(x) = x^2 + f(0) = x^2 + c.$$