This is a bit too long for a comment, so let me answer your first question.
Let $(\cdot,\cdot)$ denote the pairing of covector fields with vector fields, e.g., $(\lambda,Y) := \lambda(Y)$ for $\lambda$ a covector field and $Y$ a vector field. Think of the scalar field $(\lambda,Y)$ as the product of the covector field $\lambda$ with the vector field $Y$ and think of covariant differentation as a consistent way to define the directional differentiation of objects more complicated than scalar fields, particularly on curved spaces. Then, for all these notions of directional differentiation to be truly consistent, they should satisfy the obvious Leibniz rule with respect to taking the product of a covector field and a vector field, i.e., for any given covector field $\lambda$ and vector field $Y$,
$$
X(\lambda,Y) = (\nabla_X \lambda,Y) + (\lambda,\nabla_X Y)
$$
for any vector field $X$. What's nice is that this is actually enough to define the covariant derivative $\nabla_X \lambda$ of a covector field $\lambda$ along the vector field $X$, by defining $\nabla_X \lambda$ to be the covector field such that
$$
X(\lambda,Y) = (\nabla_X \lambda,Y) + (\lambda,\nabla_X Y),
$$
or equivalently,
$$
\nabla_X \lambda (Y) = X(\lambda(Y)) - \lambda(\nabla_X Y),
$$
for all vector fields $Y$.
Indeed, let $\Sigma \subset \mathbb{R}^3$ denote the image of $f$, let $\{e_1,e_2\}$ be your favourite frame for $T\Sigma$, i.e., a set of vector fields such that for each $p \in U$, $\{e_1(p),e_2(p)\}$ is a basis of the tangent space $T_p \Sigma \subset \mathbb{R}^3$ to $\Sigma$ at $p$, and let $\{e^1,e^2\}$ be the corresponding coframe for $T^\ast \Sigma$, i.e., a set of covector fields such that for each $p \in U$, $\{e^1(p),e^2(p)\}$ is the dual basis of the cotangent space $T^\ast_p \Sigma = (T_p \Sigma)^\ast \subset (\mathbb{R}^3)^\ast$ to $\Sigma$ at $p$ corresponding to the basis $\{e_1(p),e_2(p)\}$ of $T_p \Sigma$. Recall that any covector field $\omega$ can be uniquely written in terms of $\{e^1,e^2\}$ as
$$
\omega = (\omega,e_1)e^1 + (\omega,e_2)e^2 = \omega(e_1)e^1 + \omega(e_2)e^2.
$$
Then, we can solve Leibniz rule
$$
X(\lambda,e_j) = (\nabla_{X}\lambda,e_j) + (\lambda,\nabla_{X}e_j),
$$
i.e.,
$$
X(\lambda(e_j)) = \nabla_{X}\lambda(e_j) + \lambda(\nabla_{X} e_j),
$$
for $\nabla_{X}\lambda(e_j)$ to get
$$
\nabla_{X} \lambda(e_j) = X(\lambda(e_j)) - \lambda(\nabla_{X} e_j),
$$
which therefore forces
$$
\nabla_{X} \lambda = (X(\lambda(e_1)) - \lambda(\nabla_{X} e_1))e^1 + (X(\lambda(e_2)) - \lambda(\nabla_{X} e_2))e^2.
$$
In particular, we find that
$$
\nabla_{e_i} \lambda = (e_i(\lambda(e_1)) - \lambda(\nabla_{e_i} e_1))e^1 + (e_i(\lambda(e_2)) - \lambda(\nabla_{e_i} e_2))e^2,
$$
which is exactly what you learnt in class.
The wording "directional covariant derivative" is not widely used in the literature, but some authors (e.g. Amari, Information Geometry and Its Applications, p. 117) use it, perhaps, to distinguish from the "total covariant derivative" (see e.g. J.M.Lee, Riemannian Manifolds: An Introduction to Curvature, p. 54), which is a tensor $\nabla T$ of a higher rank, given by
$$
\nabla T (\omega_1, \dots, \omega_p, V_1, \dots, V_q, X) = \nabla_X T (\omega_1, \dots, \omega_p, V_1, \dots, V_q)
$$
As I get it, if $\nabla T$ denotes the total covariant derivative as above, then $\nabla_{X} T$ is the directional covariant derivative of $T$ in the direction of vector field $X$, and $\nabla_X T (\dots) = \nabla T (\dots, X) $.
Best Answer
Presumably, the author means that $\nabla^i$ is the pullback through $\varphi_i$ of the covariant derivative $\bar \nabla$ on $\Bbb R^n$ defined by taking the directional derivatives in the following way. If $\bar X= \sum_j \bar X^j\partial_j$ and $\bar Y = \sum_k \bar Y^k \partial_k$ are vectors fields on $\Bbb R^n$, then $\bar \nabla_{\bar X} \bar Y = \sum_{j,k} \bar X^j \frac{\partial \bar Y^k}{\partial x^j} \partial_k$.
Let $X$ and $Y$ be vector fields on $M$. The pullback covariant derivative $\nabla^i = {\varphi_i}^*\bar\nabla$ is given by the formula $\nabla^i_XY= {\varphi_i}^*\left(\bar \nabla_{{\varphi_i}_*X}{\varphi_i}_*Y\right)$, and therefore, the covariant derivative $\nabla$ on $M$ is given by
$$ \nabla_XY = \left(\sum_i\lambda_i\nabla^i\right)_XY= \sum_i\lambda_i {\varphi_i}^*\left(\bar\nabla_{{\varphi_i}_*X}{\varphi_i}_*Y\right). $$ You can check that it indeed defines a covariant derivative since the pullback $\varphi_i^*$ and the pushforward ${\varphi_i}_*$ are $\mathcal{C}^{\infty}(M;\Bbb R)$-linear.
In addition, I would like to say that this construction works on all manifolds, not only compact ones (as suggests the text OP has linked). Indeed, we only need to have a partition of unity associated to a locally finite open cover. Since all manifolds possess such a partition of unity, the same proof works.
Finally, let us mention that there is a quicker but more conceptual way to prove the existence of a covariant derivative on any manifold. By the Whitney's embedding Theorem, any manifold embeds into some euclidean space $\Bbb R^N$. The orthogonal projection of the directional derivative of vector fields on the tangent space yields such a covariant derivative. It is the famous Levi-Civita connection, which is in fact associated to the Riemannian metric induced by the embedding.
In case your question is more about the definition of the pushforward or the pullback through a diffeomorphism $\varphi\colon U\to V$, they are defined by $$ (\varphi_*X)(q) = d\varphi(\varphi^{-1}(q)) X(\varphi^{-1}(q)) $$ for $q\in V$ and $X$ a vector field on $U$, and $$ (\varphi^* Y)(p) = [d\varphi(p)]^{-1} Y(\varphi(p)) $$ for $p\in U$ and $Y$ a vector field on $V$.