Proving that Emb($D^m,N$) is homotopy equivalent to $V_m(TN)$

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I am reading online lecture notes by John Francis on h-principle. I want to prove that

Emb($D^m,N$) is homotopy equivalent to $V_m(TN)$ where $V_m(TN)$ is stiefel bundle of the tangent bundle on $N$.

So to prove this we first construct a map from $F:$Emb($D^m,N$) $\to V_m(TN)$ given by $F(f)=(f(0),df(0)(T_0(D^m)))$.

Now we construct the reverse map say $G$. So suppose we have a point of $V_m(TN) $ this is equivalent to the data a point of $N$ say $x$ and a map $i:\mathbb{R}^m \to T_xN$. Now we first put a Riemannian metric on $N$. Now scale the map $i$ such that the image of disc around the origin lies in the normal neighbourhood. Now we have a map $\lambda i:\mathbb{R}^m \to T_xN$ such that $\lambda i(D_m) \subset U$ where $U$ is a neighbourhood of $0$ such that $exp : T_xN \to N$ when restricted to $U$ is a diffeomorphism. Now the composite map $exp \circ \lambda i$ gives an embedding. This is the reverse map.

Now I want to show that this map is the homotopy inverse of $F$. For that we have to show that if $f \in$ Emb($D^m,N$) then $exp \circ df_0$ isotopic to $f$. I am unable to prove this statement. If anyone can give any hints it would be really great. Thanks.