I tried to derive the covariant derivative of a Riemannian manifold $M$ which is embedded in a higher-dimensional euclidean space $E$ (Levi-Civita connection induced from euclidean metric).
Let $\pi$ be the projection of a vector in $E$ onto the tangential space $T_pM$. Then we can define the covariant derivative of a vector field $V$ on $M$ as the euclidean partial derivative $\partial_\mu V$, projected onto the tangential space $T_pM$:
\begin{align}D_\mu V &:= \pi(\partial_\mu V)=\pi\left(\partial_\mu\left(V^\nu \vec{e_\nu}\right)\right) \\ & = \pi\left(\partial_\mu V^\nu \vec{e_\nu}+V^\nu\vec{e_{\nu,\mu}}\right) \\ \tag{1} & =\pi\left(\partial_\mu V^\nu \vec{e_\nu} + V^\nu{\Gamma^\lambda}_{\mu\nu}\vec{e_\lambda}+V^\nu K_{\mu\nu}^\lambda\vec{n_\lambda})\right) \\ & =\partial_\mu V^\nu \vec{e_\nu} + V^\nu{\Gamma^\lambda}_{\mu\nu}\vec{e_\lambda} \\ \tag{2}& =\vec{e_\nu}\left(\partial_\mu V^\nu + V^\lambda{\Gamma^\nu}_{\mu\lambda}\right) \\ & = \vec{e_\nu}\left(\delta^\nu_\lambda\partial_\mu+{\Gamma^\nu}_{\mu\lambda}\right)V^\lambda\end{align}.
In $(1)$, we just expand $\vec{e_{\nu,\mu}}=\partial_\mu\vec{e_\nu}$ into a basis of tangential vectors $\vec{e_\nu}$ and normal vectors $\vec{n_\lambda}$, where the Christoffel symbols appear naturally ($K_{\mu\nu}^\lambda$ is some sort of higher-dimensional second fundamental form, because for a hyper-surface, there is just one normal vector $\vec{n}$).
In $(2)$, we just re-label some indices.
Everything correct so far?
Now, various sources (for example this) say that the covariant derivative of a vector field should be
$$ \tag{3} D_\mu V^\nu = \partial_\mu V^\nu + {\Gamma^\nu}_{\mu\lambda}V^\lambda $$
What's the exact difference between $(2)$ and $(3)$? Why isn't it necessary to write the basis vectors $\vec{e_\mu}$ in $(3)$?
What you wrote is perfectly fine, there's no difference between the formula you derived and $(3)$. $D_\mu$, as an operator on $T_pM$, satisfies $D_\mu V = (\partial_\mu V^\nu + \Gamma^\nu_{\mu \lambda} V^\lambda) \cdot \mathbf{e}_\nu$. You can either think of $D_\mu V^\nu$ in the expression in $(3)$ as $(D_\mu V)^\nu$ as described in this comment, or write $V = V^\nu \cdot \mathbf{e}_\nu$, and identify $T_pM$ with $\Bbb R^n$ by sending $\{\mathbf{e}_\nu\}$ to the standard basis vectors and don't write the basis vectors as is standard for the Euclidean space. Either way you can justify writing it as $D_\mu V^\nu = \partial_\mu V^\nu + \Gamma_{\mu \lambda}^\nu V^\lambda$ - this is just a notation.