Let $\Phi : GL(n)\rightarrow Sym(n)$ be defiened by $\Phi (A)=AA^T$. I can not see how to get the "right" pushforward,
I.e I want help in understanding the pushforward $\Phi _*:M_I(n)\rightarrow T_I(Sym(n))$.
In terms of the tangentspace realized as derivations $\Phi _*(X)f=X(\Phi \circ f)$, how do I translate this into the present situation?
In particular how can I see that it is equal to $\Phi _*(M)=M^tM+ MM^t$
In general for a smooth map $f: M\to N$, the pushforward $(\phi_*)_x : T_xM \to T_{f(x)}N$ is given by
$$[\gamma] \mapsto [f\circ \gamma],$$
where $\gamma \in T_xM$ is a curve in $M$ so that $\gamma(0) = x$. Then in our case, if you want to calculate $(\phi_*)_I$, it is (a tangent vector $M$ is identified as $\gamma (t) =I+ tM$ $$(\phi_*)_I (M) = [ \phi(I+tM)] = [(I+tM)(I+tM)^t] = [I + t(M+ M^t) + t^2MM^t]$$
thus $(\phi_*)_I$ is given by
$$M \mapsto M+M^t$$
which is not the one you suggested. (Your map is not linear so it cannot be $(\phi_*)_I$)