If it is given that $f$ and $g$ are polynomials then the product of their degrees must be $2$ (the degree of the composition of two polynomials is the product of their individual degrees), therefore one of the two must be quadratic and the other linear. I do not think that you can assume $\deg f=2$ without loss of generality.
If $\deg f=2$ and $\deg g=1$ then $g$ must be increasing, i.e., its leading coefficient is positive. The graph of $f,$ like the graph of $f\circ g,$ is tangent to the $X$ axis (discriminant $0$) and $f\circ g$ reaches its minimum $0$ when $x=-\frac12.$ The minimum of $g\circ f$ is $g(f(-1))=1$ and, because $g$ is strictly increasing, is reached when $f(x)$ reaches its minimum $0.$ Therefore the $Y$-intercept of $g$ is 1. This also tells us that $f$ reaches its minimum at $x=-1$ and we already know that $g(-\frac12)=-1$. Therefore
Plugging this into $g(f(x))=x^2+2x+2$ gives $a=\frac14,$ and then into $f(g(x))=4x^2+4x+1$ verifies this.
Now suppose $\deg g=2$ and $\deg f=1.$ Then $f$ must increasing (positive leading coefficient). The graph of $g$, like the graph of $g\circ f,$ has minimum $1.$ The function $g\circ f$ reaches this minimum when $x=-1.$ The minimum of $f\circ g$ is $f(g(-\frac12))=0$ and, because $f$ is strictly increasing, is reached when $g(x)$ reaches its minimum $1.$ Therefore $f(1)=0.$ This also tells us that $g$ reaches its minimum at $x=-\frac12$ and we already know that $f(-1)=-\frac12.$ Therefore
Plugging this into $f(g(x))=4x^2+4x+1$ gives $a=4,$ and then into $g(f(x))=x^2+2x+2$ verifies the result.
Without any restriction on $f$ and $g$ the solutions can be very wild. If $(f_1,g_1)$ and $(f_2,g_2)$ are distinct solutions where all 4 functions leave a proper subset $A\subset\mathbb R$ and its complement intact, then two new solutions can be obtained by applying one solution to the numbers in $A$ and the other to its complement. As an example, the two polynomial solutions above map algebraic numbers to algebraic numbers and transcendental numbers to transcendental numbers.