I guess I will answer this, since I've been working on it and no one else seems too interested!
The thing I want to start with comes from the definition of a spin derivative in the original question:
$$\nabla_X^S \phi - \nabla_X^F \phi |_F = X(\phi) - X(\phi |_F) +\frac{1}{2}\sum_{1\leq i<j\leq n}\omega_{ij}s_i\cdot s_j\cdot \phi -\frac{1}{2}\sum_{1\leq i<j\leq n-1}\omega_{ij}s_i\cdot s_j\cdot \phi|_F$$
Here $F$ is a codimensional 1 subbundle of spin bundle $S$, $N$ is the normal to the submanifold $\Sigma$ and $X\in T\Sigma$. The first trick is to identify that there is an isomorphism between a $n$-dimensional spin representation and a $n-1$-dimensional representation. This comes from that link above (Baum et al 1991), but it's essential an inclusion, with the extra dimension either
$$e_{2m+1}(u^+ \oplus u^-)=(-1)^m i (u^+ - u^i)$$
or
$$e_{2m+2}(u\oplus \hat{v})=(-1)^m i (v\oplus \hat{u})$$
depending on the dimension. That isomorphism lets us set $X(\phi)=X(\phi |_F)$, so the first terms in the expression vanishes.
The next two terms mostly cancel each other - only terms in the first with $j=n$ survive. So we can write that as
$$\frac{1}{2}\sum_{1\leq i < n}\omega_{in}s_i\cdot s_n \cdot \phi$$
We need to make that look like a derivative of the normal vector times (Clifford Product) the normal vector times the spinor. Presumably we'll have to assume the normal vector is unit...so then $s_n=N$. The covariant derivative of the normal vector will be
$$ \nabla_X^M N = X(N) + N^a\nabla_X (s_a)$$
First, $X(N)=0$ since they exist in orthogonal spaces. Now (I think) the traditional definition of that covariant derivative is
$$\nabla_X (s_a)=\sum_b \omega(X)_{ab}s_b$$
Which corresponds to Baum et al when they write
$$\omega_{kl}=\langle \nabla s_k,s_l\rangle=\nabla s_{k\mu} s^{\mu}_l$$
But really this should be written $\nabla_{\nu}s_{k\mu} s^{\mu}_l$ and contracts with the components of the vector $X^{\nu}$. So the derivative of the normal vector is
$$\nabla_X^M N = N^a\sum_b \omega(X)_{ab}s_b=N^n\sum_b \omega(X)_{nb}s_b =-N^n\sum_b \omega(X)_{bn}s_b$$
using the antisymmetry of $\omega_{ab}$. Now with $N^n=1$ we have
$$\frac{1}{2}\sum_{1\leq i < n}\omega_{in}s_i\cdot s_n \cdot \phi =-\frac{1}{2} \sum_b \omega(X)_{bn}s_b \cdot N \cdot \phi = -\frac{1}{2}\nabla^M_X N \cdot N\cdot \phi$$
and we're done!
$$\nabla_X^S \phi=\nabla_X^F \phi|_F-\frac{1}{2}\nabla^M_X N \cdot N\cdot \phi$$
I will certainly take comments - it seems a bit of an awkward proof to be left to "can be verified", but maybe the authors also didn't like what it looked like.
Disclaimer: I've written everything out in painstaking detail, including explicit canonical isomorphisms, which definitely obscures the proof - but the point for me is to make it clear and explicit once and for all to avoid confusion later on.
The entire argument can be summarised by the following commutative diagram:
which arrises by looking locally $(U\subset M)$. The point to note is that the map
$$
\Gamma(U\times\mathbb R^r)\to\Omega^0_\rho(U\times\text{GL}_r\mathbb R,\mathbb R^r)
$$
in the diagram is the pull-back map $\pi^*$, and that the map
$$
\Omega^1_\rho(U\times\text{GL}_r\mathbb R,\mathbb R^r)\to \Omega^1(U)\otimes\Gamma(U\times\mathbb R^r)
$$
in the diagram is the pull-back $\sigma^*$ (where $\sigma\in\Gamma(\text{Fr}E\vert_U)$ is a local section we arrive at by choosing a local trivialisation).
Remark: Initially I wrote the text below without the diagram, and in the end I realised which diagram belongs to my proof. Since the diagram is very clear, I've put it first, but now comes the original text-based proof:
Choose a local trivialisation $\phi:E\vert_U\stackrel{\sim}{\to} U\times\mathbb R^r$, and write $\sigma_i=\Gamma(\phi)^{-1}(e_i)$ (where $\Gamma$ is the section functor). We have an induced trivialisation
$$
\Phi:\text{Fr}E\vert_U\stackrel{\sim}{\to} U\times\text{GL}_r\mathbb R,
$$
and we let $\sigma=\Gamma(\Phi)^{-1}(I)$, which means fiber-wise that $\sigma_p:e_i\mapsto\sigma_i\vert_p$. Our trivialisation induces an isomorphism between vector bundles:
$$
\text{Fr}E\vert_U\times_1\mathbb R^r\simeq(U\times\text{GL}_r\mathbb R)\times_1\mathbb R^r\cong U\times\mathbb R^r.
$$
Under these identifications, the inverse of the isomorphism in Theorem 31.9 is locally given given by the pull-back of the projection map $\pi^*$:
\begin{multline*}
\pi^*:\Omega^k(U)\otimes\Gamma(U\times\mathbb R^k)\stackrel{1\otimes\Gamma\phi^{-1}}{\longrightarrow}\Omega^k(U)\otimes\Gamma(\text{Fr}E\vert_U\times_1\mathbb R^k)\\
\stackrel{\pi^*}{\to}\Omega^k(\text{Fr}E\vert_U)\otimes\Gamma(\pi^*(\text{Fr}E\vert_U\times_1\mathbb R^k))\to
\Omega^k(\text{Fr}E\vert_U)\otimes\Gamma(\text{Fr}E\vert_U\times\mathbb R^k).
\end{multline*}
We can verify explicitly that the map above on the sections sends $e_i$ indeed to its pull-back $\pi^*e_i=e_i\in\Gamma(U\times\mathbb R^r)$:
\begin{gather*}
\Gamma(U\times\mathbb R^r)\simeq\Gamma(\text{Fr}E\vert_U\times_1\mathbb R^r)\to\Gamma(\pi^*(\text{Fr}E\vert_U\times_1\mathbb R^r))\cong\Gamma(\text{Fr}E\vert_U\times\mathbb R^r)\\
(p\mapsto (p,e_i))\mapsto (p\mapsto [\sigma_p,e_i])\mapsto (\tau_p\mapsto (\tau_p,[\sigma_p,e_i]))\mapsto(\tau_p\mapsto (\tau_p,e_i)).
\end{gather*}
Write
$$
\omega=\sum\tilde\omega^i_j\otimes e^i_j\in\Omega^1(\text{Fr}E\vert_U)\otimes\Gamma(\text{Fr}E\vert_U\times\mathfrak{gl}_r\mathbb R),
$$
for the restricted connection form $\text{Fr}E\vert_U$, and write
$$
\omega_\sigma=\sum\omega^i_j\otimes e^i_j\in\Omega^1(U)\otimes\Gamma(U\times\mathfrak{gl}_r\mathbb R)
$$
for the local connection form of relative to the frame $\sigma$, meaning that $\nabla\sigma_i=\omega^j_i\sigma_j$. Since $\sigma^*\omega=\omega_\sigma$ we have $\sigma^*\tilde\omega^i_j=\omega^i_j$. Note that
$$
\omega\cdot e_i=\tilde\omega^j_i e_j,
$$
while $d(e_i)=0$, so
$$
D(e_i)=d(e_i)+\omega\cdot e_i=\tilde\omega^j_i e_j
$$
To go back to $\Omega^1(U)\otimes\Gamma(U\times\mathbb R^r)$, we simply need to pull back along $\sigma$ to obtain $\sigma^*\omega^j_i e_j=\omega^j_i e_j$; because $\sigma$ corresponds to the trivial section of $\text{Fr}E\vert_U\simeq(U\times\text{GL}_r\mathbb R))$ and by the naturality of $\omega\mapsto\omega^\flat$ we may apply the lemma below.
Lemma.
Let $P=M\times G\to M$ be a trivial principal $G$ bundle, $\rho:G\to\text{GL}(V)$ a representation. Denote by $\phi:P\times_\rho V\cong M\times V,[(x,g),v]\mapsto (x,v)$ the identification with the trivial bundle. Let $\sigma:M\to P,m\mapsto (m,e)$ be the canonical section of the trivial bundle. Then the isomorphism $\omega\mapsto\omega^\flat$ in Theorem 31.9 under our identification with the trivial bundle is given by the pull-back under $\sigma$:
$$
\sigma^*:\Omega^k_\rho(P,P\times V)\stackrel{(-)^\flat}{\to} \Omega^k(M)\otimes\Gamma(P\times_\rho V)\stackrel{\phi}{\to}\Omega^k(M)\otimes\Gamma(M\times V).
$$
Proof.
Let $\omega\in\Omega^k_\rho(P,P\times V)$. Write
$$
\omega_p(v_1,\dots,v_k)=(p,\overline\omega_p(v_1,\dots,v_k))\in\{p\}\times V.
$$
Then
\begin{align*}
\phi\omega^\flat_x(v_1,\dots,v_k)&=\phi[\omega_{\sigma(x)}(T_x\sigma v_1,\dots,T_x\sigma v_k)]=\phi[\omega_{(x,e)}(T_x\sigma v_1,\dots, T_x\sigma v_k)]\\
&=\phi[(x,e),\overline\omega_{(x,e)}(T_x\sigma v_1,\dots, T_x\sigma v_k)]=(x,\sigma^*\overline\omega_x(v_1,\dots,v_k)).
\end{align*}
$\tag{$\square$}$
So we see that our induced map sends $\sigma_i$ to $\omega^j_i\sigma_j=\nabla\sigma_i$. Since we also have the Leibniz condition, it follows that our induced map corresponds to the original connection $\nabla$.
Best Answer
Ok, let $X\in \mathfrak{X}(M)$, $F \in \mathscr{T}^1_1(M)$, and $\nabla$ be any affine connection on $TM$. I'll use Einstein's summation convention and local coordinates.
The idea here is that $$\begin{split}(\nabla_XF)^i_j &= {\rm d}x^{i}((\nabla_XF)(\partial_j)) \\ &= {\rm d}x^{i}(\nabla_X(F(\partial_j)) - F(\nabla_X\partial_j)) \\ &= {\rm d}x^{i}(\nabla_X(F_j^k\partial_k) - F(\nabla_X\partial_j)) \\ &= {\rm d}x^i(X(F^k_j)\partial_k + F^k_j\nabla_X\partial_k - F(\nabla_X\partial_j)) \\ &= X(F^i_j) + {\color{red}{{\rm d}x^{i}(F^k_j\nabla_X\partial_k - F(\nabla_X\partial_j))}}.\end{split}$$We are done once we show that the term in red vanishes when subjected to contraction. On one hand, we have that $$F^k_j\nabla_X\partial_k = F^k_jX^\ell\nabla_{\partial_\ell}\partial_k = F^k_jX^\ell \varGamma_{\ell k}^i\partial_i,$$and on the other $$F(\nabla_X\partial_j) = F(X^\ell\nabla_{\partial_\ell}\partial_j )=F(X^\ell \varGamma_{\ell j}^{k}\partial_k) = F_k^iX^\ell \varGamma_{\ell j}^k\partial_i,$$so that $${\color{red}{{\rm d}x^{i}(F^k_j\nabla_X\partial_k - F(\nabla_X\partial_j))}} = F^k_jX^\ell\varGamma_{\ell k}^i - F^i_{k}X^\ell\varGamma_{\ell j}^k = X^\ell(F^k_j\varGamma_{\ell k}^i - F^i_k\varGamma_{\ell j}^k).$$The factor $X^\ell$ is irrelevant, so make $i=j$ on $F^k_j\varGamma_{\ell k}^i - F^i_k\varGamma_{\ell j}^k$ to obtain $$F^k_i\varGamma_{\ell k}^i - F^i_k\varGamma_{\ell i}^k = 0,$$since the second term is obtained from the first one by renaming the dummy indices $i\leftrightarrow k$.