No, that's too much to ask for. If you take any convex cone in $\mathbb R^n$ (any such cone is both $CAT(0)$ and Alexandrov of $curv\ge 0$) then the tangent space at the origin is just the cone itself. So it need not be any kind of quotient of $\mathbb R^n$. The above picture is general. If an $n$-dimensional space is both locally $CAT(0)$ and Alexandrov of $curv\ge 0$ then the exponential map at any point is an isometry on a small ball and the tangent cone is a convex cone in $\mathbb R^n$.

No, that's too much to ask for. If you take any convex cone in $\mathbb R^n$ (any such cone is both $CAT(0)$ and Alexandrov of $curv\ge 0$) then the tangent space at the origin is just the cone itself. So it need not be any kind of quotient of $\mathbb R^n$. The above picture is general. If an $n$-dimensional space is both locally $CAT(0)$ and Alexandrov of $curv\ge 0$ then the exponential map at any point is an isometry on a small ball and the tangent cone is a convex cone in $\mathbb R^n$.

BTW, regarding Igor's comment about local extendability of geodesics and a result of Nikolaev, in a recent paper with Kell and Ketterer we showed that if a space $X$ is both locally $CAT(K)$ and $CD(k,n)$ (has generalized Ricci curvature bounded below, which is a weaker condition than Alexandrov curvature bounded below) then $X$ is a manifold with boundary and the interior points are exactly the points where all geodesics locally extend.

No, that's too much to ask for. If you take any convex cone in $\mathbb R^n$ (any such cone is both $CAT(0)$ and Alexandrov of $curv\ge 0$) then the tangent space at the origin is just the cone itself. So it need not be any kind of quotient of $\mathbb R^n$. The above picture is general. If an $n$-dimensional space is both $CAT(0)$ and Alexandrov of $curv\ge 0$ then the exponential map at any point is an isometry on a small ball and the tangent cone is a convex cone in $\mathbb R^n$.

No, that's too much to ask for. If you take any convex cone in $\mathbb R^n$ (any such cone is both $CAT(0)$ and Alexandrov of $curv\ge 0$) then the tangent space at the origin is just the cone itself. So it need not be any kind of quotient of $\mathbb R^n$. The above picture is general. If an $n$-dimensional space is both locally $CAT(0)$ and Alexandrov of $curv\ge 0$ then the exponential map at any point is an isometry on a small ball and the tangent cone is a convex cone in $\mathbb R^n$.