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Oct 8, 2021 at 12:13 comment added Bad English In order to make my answer not completely useless let me mention a counter-example to your original question, where the projection definitely gives an isomoprhism. Take $S=\mathbb{P}^1$. Then tangent vectors to $F_x$ ($\tau_r$) are determined by some $r\in Hom(\nu_{S/M},T_S)=\mathbb{C}$. Considering a diagonal action of $SL(2)$ on $M$ we obtain an action on $\mathbb{C}$ which is trivial, hence $\tau_r$ is invariant under the action and at least one of two projections is isomorphism.
Oct 8, 2021 at 12:09 comment added Bad English Let $\phi:Tot(\nu_{S/M})\to Tube(S/M)$ be any tubular neighborhood. Consider a "retraction" fibers $F_x=\phi^{-1}(x)$ for $x\in S$. Tangent vectors along $F_x$ define a vector bundle $\tau_r\subset {T_M}|_{S}=\nu_{S/M}\oplus T_S$, which is transversal to $T_S$. In other words $\tau_r$ is a deformation of $\nu_{S/M}$ defined by an element in $r\in Hom(\nu_{S/M},T_S)$. If one can produce an automorphism $\alpha$ of $Tot(\nu_{S/M})$ which differential maps $\tau_r$ to $\nu_{S/M}$, then we are done. I don't see how one can do this.
Oct 8, 2021 at 12:08 comment added Bad English @SimonParker, oops, you absolutely right. I unintentionally took the retraction into account and impose stronger conditions on the definition of a tubular neighborhood. Namely, if the tubular neighborhood morphism restricted to $S$ maps tangent vectors along the normal bundle to tangent vectors along the retraction fibers, then the projection you mentioned above gives an isomorphism and we can obtain a connection.
Oct 6, 2021 at 17:26 comment added Simon Parker Thanks for your answer. I do not understand the first step of the definition of the connection, namely: "The tubular neighborhood for each $x\in S$ defines identification $f_x$ of a neighborhood $0\in T_x$ with a neighborhood of $x\in S$". The tubular neighbourhood gives an embedding of a neighborhood of $0$ in $T_xS$ in $S \times S$, so presumably you want to compose this with one of the projections to get the desired identification. But it is easy to find examples where this composition is not an isomorphism.
Oct 5, 2021 at 17:48 history answered Bad English CC BY-SA 4.0