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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 $\mathbb R^n$.