The first map you wrote down isn't a ring homomorphism (already it doesn't send $1$ to $1$), so it doesn't have an algebro-geometric interpretation. The obvious ring homomorphism goes in the other direction, namely what you might call the "Chinese remainder theorem map"
$$A[x]/(fg) \ni p \mapsto (p \bmod f, p \bmod g) \in A[x]/f \times A[x]/g$$
given by the two obvious projections. Now there is a corresponding morphism of affine schemes going in the other direction from what you wrote, namely the canonical map $Z(f) \sqcup Z(g) \to Z(fg)$.
This map does not seem to me to have any obvious relationship to the Sylvester matrix. In the standard treatment of resultants you will see the domain of the Sylvester map being given as polynomials of degrees $< \deg f$ and $< \deg g$, rather than as elements of the quotients $A[x]/f, A[x]/g$; note that these quotients are not free $A$-modules on the monomial basis unless the leading coefficients of $f$ and $g$ are invertible over $A$.
However in a suitable basis the determinant of the Chinese remainder map is also the resultant (up to scale or something), since this determinant shares the same property the resultant does that it vanishes iff the polynomials have a common root, and the resultant as a polynomial is known to be irreducible. So this leads to an alternative definition of the resultant.
Some details. It seems fairly annoying to write down the Chinese remainder map in the monomial basis, but we can use the following alternative basis: if $\deg f = n$ and $\deg g = m$, and assuming WLOG that $n \le m$, we can use the basis $\{ 1, x, x^2, \dots x^{n-1}, f, fx, fx^2, \dots fx^{m-1} \}$ of $A[x]/(fg)$ (assuming for simplicity that the leading coefficient of $f$ is invertible over $A$). Then the map to $A[x]/f$ is easy to describe since it just sends this basis to $1, x, x^2, \dots x^{n-1}, 0, 0, \dots$. The map to $A[x]/g$ also just sends $\{1, x, x^2, \dots x^{n-1} \}$ to those same monomials but the later terms depend on the coefficients of $fx^k \bmod g$. The resulting $(n+m) \times (n+m)$ matrix has an upper left block which is just the identity matrix so its determinant ends up being the determinant of the $m \times m$ lower right block, consisting of the coefficients of $fx^k \bmod g$. The claim that this recovers the resultant appears as Theorem 1.18 here but I don't know if it has a name or who it's due to. It may be related to the Bezout matrix.
Here's an explicit calculation for $n = m = 2$ with monic polynomials. If we write $f(x) = x^2 + f_1 x + f_0$ and $g(x) = x^2 + g_1 x + g_0$ then the ordinary resultant coming from the Sylvester matrix is
$$\begin{eqnarray*} \text{Res}(f, g) &=& \det \left[ \begin{array}{cc} 1 & f_1 & f_0 & 0 \\ 0 & 1 & f_1 & f_0 \\ 1 & g_1 & g_0 & 0 \\ 0 & 1 & g_1 & g_0 \end{array} \right] \\
&=& (f_1 - g_1)(f_1 g_0 - f_0 g_1) + (f_0 - g_0)^2. \end{eqnarray*}$$
The Chinese remainder map, on the other hand, with respect to the bases $\{ 1, x, f, fx \}, \{ 1, x \}, \{ 1, x \}$ as above, is given by
$$\left[ \begin{array}{cc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & f_0 - g_0 & -(f_1 - g_1) g_0 \\ 0 & 1 & f_1 - g_1 & -(f_1 - g_1) g_1 + (f_0 - g_0) \end{array} \right]$$
and its determinant reduces to a $2 \times 2$ determinant which, after some simplification, coincides with the resultant. If we allowed leading coefficients there would be some denominators, I guess.
