First, perhaps you misread the requirement: it is not itself $M$ that must have codimension 1, it is instead the boundary of $M$, denoted $\partial M$, that must have codimension 1. But the phrase "... with boundary of codimension 1... " leaves out some information which could perhaps clarify the situation: that phrase should be parsed as
... such that the boundary of $M$ has codimension in $\mathbb R^d$ equal to 1...
Equivalently, $M$ itself must have codimension in $\mathbb R^d$ equal to 0.
So, why must $\partial M$ have codimension $1$ in $\mathbb R^d$?
As soon as you write the words "$\nu$ is the normal field and $V_{\partial M}$ is the projection of $V$ onto the tangent space", one wonders: normal field of what? Tangent space of what? The only sensible answer that I can see is that $\nu$ is the normal field to $\partial M$ and $V_{\partial M}$ is the projection of $V$ onto the tangent space of $\partial M$.
And in order that $\partial M$ even possess a normal field, it must have codimension 1. So, in order for the $v$ term in equation (1) to even be defined, $\partial M$ must have codimension 1.
(It might be clearer if one rewrites equation (1) with proper quantifiers: each term should have the argument $x$ in the correct position, sort of like you did in the later equations; and the equation should hold for each $x \in \partial M$.)
The point here is that a submanifold of dimension $n \ge 2$ or higher does not have a well-defined normal field. It does have a well-defined normal bundle, which is a vector bundle of dimension $n$. For example, for a circle $C$ embedded in $\mathbb R^3$, which has codimension 2, its normal bundle has fibers of dimension $2$: at each point $x \in C$, the normal plane $N_x C$ is the 2-dimensional subspace of $T_x \mathbb R^3$ ($= \mathbb R^3$) which is normal to the 1-dimensional tangent line $T_x C$.
In general, for an codimension $m$ submanifold $B \subset \mathbb R^n$, the normal bundle is an $m$-dimensional vector bundle over $B$, whose fiber $N_x B$ is the $m$-dimensional subspace of $T_x \mathbb R^n$ (which is identified with $\approx \mathbb R^n$) that is normal to the $n-m$ dimensional subspace $T_x B$ of $T_x \mathbb R^n$. It follows that there is an orthogonal direct sum
$$T_x \mathbb R^n = T_x B \oplus N_x B
$$
And then, as you asked in the comments, for the case that $n=d$ and that $B = \partial M$ has codimension 1 in $\mathbb R^n$, one obtains
$$T_x \mathbb R^d = T_x (\partial M) \oplus N_x (\partial M)
$$
Your expression is wrong. There are no derivatives of $\omega$ involved (also notice that there is no first derivative of $f$ in the last parcel of your formula). Recall that for a smooth local frame $\{e_1, \cdots, e_n\}$ the connection forms $\{\omega_{j}^{k} \}_{1 \leq k \leq n}$ are determined by $$\nabla_{e_i} e_j = \sum_{1 \leq k \leq n} \omega_{j}^{k}(e_i) e_k$$
Or, equivalently, $$\nabla e_j = \sum_{1 \leq k \leq n} \omega_{j}^{k}(\bullet) e_k$$
which is usually written as:
$$\nabla e_j = \sum_{1 \leq k \leq n} \omega_{j}^{k} e_k$$
In the case that the local frame is a coordinate frame (which is apparently the case you want), by definition
$$\omega_j^k(\partial_{i}) = \Gamma_{ij}^k$$
Now, since $\nabla^2 f$ is a $(0, 2)$ tensor and $\{ \mathrm{d}x^i \otimes \mathrm{d}x^j \} $ forms a basis of the space of such tensors, we have:
$$\nabla^2 f = \sum_{1 \leq i, j \leq n} (\nabla^2 f)_{ij} \mathrm{d}x^i \otimes \mathrm{d}x^j $$
But
\begin{aligned}
(\nabla^2 f)_{ij} = \nabla^2_{\partial_i, \partial_j} f &= \nabla_{\partial_i} \nabla_{\partial_j} f - (\nabla_{\partial_i} \partial_j)f \\
&= \frac{\partial^2 f}{\partial x^i \partial x^j} - \sum_{1 \leq k \leq n} (\Gamma_{ij}^k \partial_k) f \\
&= \frac{\partial^2 f}{\partial x^i \partial x^j} - \sum_{1 \leq k \leq n} \Gamma_{ij}^k \frac{\partial f}{\partial x^k}\\
\end{aligned}
Which can also be written as
$$(\nabla^2 f)_{ij} = \frac{\partial^2 f}{\partial x^i \partial x^j} - \sum_{1 \leq k \leq n} \omega_{j}^k(\partial_i) \frac{\partial f}{\partial x^k}$$
So we have:
$$\nabla^2 f = \sum_{1 \leq i, j \leq n }\left( \frac{\partial^2 f}{\partial x^i \partial x^j}- \sum_{1 \leq k \leq n} \omega_{j}^k(\partial_i) \frac{\partial f}{\partial x^k}\right) \mathrm{d}x^i \otimes \mathrm{d}x^j $$
Clearly,
$$\sum_{1 \leq i \leq n} \omega_{j}^k(\partial_i) \mathrm{d} x^i = \omega_{j}^k$$
Therefore:
$$\nabla^2 f = \left( \sum_{1 \leq i, j \leq n } \frac{\partial^2 f}{\partial x^i \partial x^j} \mathrm{d}x^i \otimes \mathrm{d}x^j \right) - \sum_{1 \leq j, k \leq n} \frac{\partial f }{\partial x^k} \omega_{j}^k \otimes \mathrm{d}x^j$$
which can also be written as
$$\nabla^2 f = \left( \sum_{1 \leq i, j \leq n } \frac{\partial^2 f}{\partial x^i \partial x^j} \mathrm{d}x^i \otimes \mathrm{d}x^j \right) - \sum_{1 \leq j, k \leq n} \left(\frac{\partial }{\partial x^k} \otimes \omega_{j}^k \right)(f, \bullet) \otimes \mathrm{d}x^j$$
REMARK: A vector field $X$ can be seen as a $(1, 0)$ tensor field, which acts on the space of $1$-forms $\Gamma(T^{*} M)$ by $X(\omega) = \omega(X)$. $\nabla X$ can be seen as a $(1, 1)$ tensor field: either by seeing it as a map which takes a vector field as input and outputs another vector field, os a map which takes as input a real smooth function and another vector field and outputs a real smooth function, that is:
\begin{aligned}
\nabla X: \Gamma(T^{*} M) \times \Gamma(T M) &\to \mathcal{C}^{\infty}(M) \\
(\beta, Y) &\mapsto (\nabla_Y X)(\beta) = \beta(\nabla_Y X)
\end{aligned}
Now, if $V$ is a vector field and $\beta$ is a $1$-form, we have the tensor product $\beta \otimes V: \Gamma(T M) \times \Gamma(T^{*} M) \to \mathcal{C}^{\infty}(M)$, which is a tensor field of type $(1, 1)$ given by $$(\beta \otimes V)(Z, \alpha) = \beta(Z)V(\alpha)= \beta(Z) \alpha(V)$$
In particular, $\nabla e_i$ is a $(1, 1)$ tensor field given by $$(\nabla e_i)(X, \alpha) = (\nabla_X e_i)(\alpha) = \sum_{1 \leq j \leq n} (\omega_i^j(X) e_j)(\alpha) = \sum_{1 \leq j \leq n} \omega_i^j(X) e_j(\alpha) = \sum_{1 \leq j \leq n} (\omega_i^j \otimes e_j)(X, \alpha)$$
Since $X$ and $\alpha$ are arbitrary, we can write:
$$\nabla e_i = \sum_{1 \leq j \leq n} \omega_i^j \otimes e_j$$
Best Answer
Note that for $F\colon M \to\mathbb{R}^d$, we have the differential map $dF_p\colon T_pM\to T_{F(p)}\mathbb{R}^d$ for all $p\in M$, i.e. $dF_p$ transports tangential vectors in $T_pM$ to tangential vectors in $T_{F(p)}\mathbb{R}^d$. The tangential vector they transport here via the differential map is the gradient of a function $f$ on $M$. So the given notation means $$ (\nabla^M f)_p := dF_p((\nabla f)_p)$$ for all points $p\in M$, where they denote $dF_p((\nabla f)_p)$ with respect to local coordinates.
Looking at an example like the embedding of a 2-sphere into $\mathbb{R}^3$ is certainly a good way to get a better understanding of the definition. Using standard coordinates on $\mathbb{S}^2$, an embedding can (almost everywhere) be given by $$ F\colon [0,\pi]\times[0,2\pi)\to\mathbb{R}^3,\quad (\theta,\phi)\mapsto\left(\begin{array}{c}\sin\theta\cos\phi \\ \sin\theta\sin\phi\\ \cos\theta\end{array}\right), $$ which yields $$ dF = \left(\frac{\partial F}{\partial x_i}\right)_{x_1 = \theta, x_2 = \phi} = \left(\begin{array}{cc}\cos\theta\cos\phi & -\sin\theta\sin\phi \\ \cos\theta\sin\phi & \sin\theta\cos\phi\\ -\sin\theta & 0\end{array}\right).$$ Then, with respect to these coordinates, you can compute the gradient vector $\nabla^{\mathbb{S}^2}f$ in $\mathbb{R}^3$, which is tangential to the sphere embedded into $\mathbb{R}^3$, from the gradient vector $\nabla f$ on $\mathbb{S}^2$ by multiplication (for any differentiable function $f$): $$ \nabla^{\mathbb{S}^2}f = DF\cdot\nabla f = \left(\begin{array}{cc}\cos\theta\cos\phi & -\sin\theta\sin\phi \\ \cos\theta\sin\phi & \sin\theta\cos\phi\\ -\sin\theta & 0\end{array}\right)\cdot \nabla f. $$ Simple example: $f=\phi$ gives you $ \nabla^i \phi = g^{ij}(df)_j $, hence $$ \nabla\phi = \left(\begin{array}{cc} 1 & 0 \\ 0 & \frac{1}{\sin^2\theta}\end{array}\right)\left(\begin{array}{c} 0 \\ 1 \end{array}\right) = \left(\begin{array}{c} 0 \\ \frac{1}{\sin^2\theta}\end{array}\right) $$ and thus $$ \nabla^{\mathbb{S}^2}\phi = \left(\begin{array}{c} -\frac{\sin\phi}{\sin\theta} \\ \frac{\cos\phi}{\sin\theta}\\0\end{array}\right) $$ tangential to $\mathbb{S}^2$ in $\mathbb{R}^3$.