I think one way of explaining this is via quadratic forms. The usual correspondence sends $X$ to the quadratic form $X(v)=vXv^t,$ $v$ a row vector. But in characteristic $2$, this formula simplifies to

$$X(v)=(vX_0)^2.$$

Now examine what happens to $AXA^t$. For any characteristic, we get $AXA^t(v)=vAXA^tv^t=(vA)X(vA)^t=X(vA).$ In characteristic $2$, this implies that $$(v(AXA^t)_0)^2=(vA(X_0))^2,$$

so

$$(AXA^t)_0=A(X_0)$$

as desired.

I think one way of explaining this is via quadratic forms. The usual correspondence sends $X$ to the quadratic form $X(v)=vXv^t,$ $v$ a row vector. But in characteristic $2$, this formula simplifies to

$$X(v)=(vX_0)^2.$$

Now examine what happens to $AXA^t$. For any characteristic, we get $AXA^t(v)=vAXA^tv^t=(vA)X(vA)^t=X(vA).$ In characteristic $2$, this implies that $$(v(AXA^t)_0)^2=(vA(X_0))^2,$$

so

$$(AXA^t)_0=A(X_0)$$

as desired.

You can also phrase this in a representation-theoretic language (following the suggestion of user44191). Take the algebraic group $\operatorname{GL}(n)$ over $\mathbb{Z}.$ It has representations $U$ and $V$ (again over $\mathbb{Z}$) corresponding to $n\times n$ symmetric matrices and quadratic forms on $\mathbb{Z}^n.$ The natural map $U\rightarrow V$ is an isomorphism once you localize away from $2$, but let us examine the map $U/2U\rightarrow V/2V.$ This map is our map $X\mapsto(vX_0)^2,$ which shows that $U/2U$ has the desired quotient.