Assume $F: M \times [0,T) \rightarrow \mathbb R^{n+1}$ is solution of some flow like $$ (\partial_t F(q,t))^\bot= H(F(q,t)) ~~~(q,t)\in M \times [0,T) $$ $M$ is a n-dim smooth Rie-manifold. $H$ is mean curvature. In fact, this flow is equal to mean curvature flow up to tangential diffeomorphisms. For proving this , we need a diffeomorphism $\varphi : M \times[0 ,T )\rightarrow M $ satisfy $$ \nabla _q F(\varphi(p,t),t)(\partial _t \varphi(p,t))=-(\partial_t F(q,t))^\top $$ But I can't see why $\varphi$ always existence ? $\nabla$ is Euclidean connection.
This question origin from 82 page of Mean Curvature Flow of Cylindrical Graphs
I don't understand the notation in the post, nor the relation to connections. However, here is how you can construct the desired reparametrization.
Let $F:M\times[0,t)\to\mathbb{R}^{n+1}$ satisfy your condition. That is, $$(\partial_t F(q,t))^\bot= H(F(q,t)) ~~~(q,t)\in M \times [0,T).$$ Then there is a time-dependent vector field $X_t\in\mathfrak{X}(M),\;t\in[0,T),$ satisfying $$ H(F(q,t))-\partial_tF(q,t)=d(F_t)_q(X_t(q)). $$ Let $\varphi:M\times[0,T)\to M$ be the flow of $X_t$. Then, it follows from the chain rule that the reparametrized family of embeddings $F'_t:=F_t\circ\varphi_t$ is mean curvature flow.
It should be noted that the flow of the vector field $X_t$ is not necessarily defined on all $M\times[0,T)$. This issue can be resolved, for example, by requiring $M$ to be compact.