Pertaining to sectional curvature:
$f: M \to M$ is a $C^1$-diffeomorphic map on Riemannian manifold $M$, $K_p(u,v)$ is the sectional curvvature at $p \in M$ and $u,v \in T_pM$ independent. Want to see if the following hold:
$$K_{f(p)}(Df(u), Df(v)) \le C_p \cdot K_p(u,v),$$
where the constant in $ C_p$ only depends on $p$?
No: $K_p(u,v)$ could be $0$ while $K_{f(p)}(f_*u,f_*v)$ is positive (for instance, $M$ could be a surface that is flat in some parts and has positive curvature in other parts). More generally, there is nothing you can say relating $K_p(u,v)$ and $K_{f(p)}(f_*u,f_*v)$ that you cannot say relating sectional curvature functions at any points of any manifolds; that is, $f$ is totally irrelevant to your question. To be precise, given two $n$-manifolds $M_0$ and $M_1$, points $p\in M_0$ and $q\in M_1$, and a linear isomorphism $g:T_pM_0\to T_qM_1$, you can take small neighborhoods $U$ of $p$ and $V$ of $q$ and a diffeomorphism $f:U\cap V\to U\cap V$ which swaps $U$ and $V$, maps $p$ to $q$, and maps $T_pM_0\to T_qM_1$ via $g$. You then have $K_p(u,v)=K_q(g(u),g(v))$ for all $u$ and $v$.
(On the other hand, if you assume that $K_p$ is never $0$, then trivially such a $C_p$ exists since $K_{f(p)}$ is bounded and $K_p$ is bounded away from $0$ and always has the same sign, since for instance it suffices to consider the case where $u$ and $v$ are orthonormal and the space of unordered pairs of orthonormal tangent vectors $u$ and $v$ is compact and connected. Or, to use a bit more machinery, you can think of $K_p$ and $K_{f(p)}$ as continuous functions on a Grassmannian of $2$-planes, which is compact and connected.)