Here is my counterexample: Take $f(x)=\|x\|^4$, then for any $x\neq 0$, by spherical symetry, $x$ is a eigenvector of $Hf|_x$. we have then $$Hf|_x(x)=a\|x\|^2 x $$ We can choose $x=v$ an eigenvector on $\frac{1}{2}(A+A^T)$. ie $\frac{1}{2}(A+A^T)(v)=\lambda v$ then with $t\in \mathbb{R}$ $$ (Hg)|_{tv}=(Hf+A)|_{tv}(v)=(a\|v\|^2 t^2 +\lambda) v $$ Therefore for $t^2=-\lambda/(a\|v \|^2)$. $Hg|_{tv}(v)=0$ and it is then not invertible.

Here is my counterexample: Take $f(x)=\|x\|^4$, then for any $x\neq 0$, by spherical symetry, $x$ is a eigenvector of $Hf|_x$. we have then $$Hf|_x(x)=a\|x\|^2 x $$ We can choose $x=v$ an eigenvector on $\frac{1}{2}(A+A^T)$. ie $\frac{1}{2}(A+A^T)(v)=\lambda v$ then with $t\in \mathbb{R}$ $$ (Hg)|_{tv}=(Hf+A)|_{tv}(v)=(a\|v\|^2 t^2 +\lambda) v $$ Therefore for $t^2=-\lambda/(a\|v \|^2)$. $Hg|_{tv}(v)=0$ and it is then not invertible.

For a more general counterexample, take $f$ such that there exist $x_1,x_2$ with $det(Hf(x_1))<0$ and $det(Hf(x_2))>0$, then for comparatively small random matrix $A$, $det(Hf(x_2)+A)>0$ and $det(Hf(x_1)+A)<0$ and by continuity. There exist $x_0$ such that $Hg(x_0)$ is not invertible.

Here is my counterexample: Take $f(x)=\|x\|^4$, then for any $x\neq 0$, by spherical symetry, $x$ is a eigenvector of $Hf|_x$. We can choose $x=v$ an eigenvector on $\frac{1}{2}(A+A^T)$ with negatif eigenvalue and study then $Hg|_{\lambda x}$ for $\lambda\in [0,\infty[$. By continuity there exists $\lambda$ such that $Hg|_{\lambda x}(x)=0$

Here is my counterexample: Take $f(x)=\|x\|^4$, then for any $x\neq 0$, by spherical symetry, $x$ is a eigenvector of $Hf|_x$. we have then $$Hf|_x(x)=a\|x\|^2 x $$ We can choose $x=v$ an eigenvector on $\frac{1}{2}(A+A^T)$. ie $\frac{1}{2}(A+A^T)(v)=\lambda v$ then with $t\in \mathbb{R}$ $$ (Hg)|_{tv}=(Hf+A)|_{tv}(v)=(a\|v\|^2 t^2 +\lambda) v $$ Therefore for $t^2=-\lambda/(a\|v \|^2)$. $Hg|_{tv}(v)=0$ and it is then not invertible.