Definition for Covariant Derivative

There is a very intuitive way to understand the covariant derivative (for the Levi-Civita connection) for the special case of isometrically embedded submanifolds in $\mathbb R^n$. Roughly speaking, first take the usual derivative in $\mathbb R^n$, and then project the answer onto the tangent plane of the submanifold. The image of the projection is then the covariant derivative.

In fact, this is how the concept of covariant derivative if often introduced to undergraduates in their first differential geometry course. For instance, if you consider the sphere $S^2 \subset \mathbb R^3$, then it is easy (both analytically and visually) to prove that the tangent vector field to a great circle has zero covariant derivative and therefore that great circles must be geodesics.

One can motivate the covariant differentiation using only vector calculus. It works for an oversimplified case though (but since the OP doesn't accept either the definition via Ehresmann connection nor the vector bundle definition, may be it's justified.)

Consider new coordinates $y^i$ on $\mathbb{R}^n$ (e.g. spherical). We require orthonormality on that coordinate system, $$\mathrm{d}s^2=(u_i)^2\mathrm{d}y^i.$$ One has $\mathbf{x}=x^i(y)e_i$ and defines $$e'_j=\frac{\partial\mathbf{x}(y) }{\partial y^j}.$$ Then the metric is given by $$g_{ij}=e_i'\cdot e_j'.$$ If one considers a vector field $X:\mathbb{R}^n\to \mathbb{R}^n$ one can write $X=X^ie_i'$. If we now wish to differentiate $X$, we have to take into account the change of the components $X^i$ and of the basis $e'_j$, which is no longer rigid. That is $$\frac{\partial X }{\partial y^j}=\frac{\partial (X^i e'_i)}{\partial y^j}=\frac{\partial X^i}{\partial y^j}e_i'+X^i\frac{\partial e'_i}{\partial y^j}.$$ One can write $\frac{\partial e'_i}{\partial y^j}$ as linear combination of the $e_i'$ s, i.e. for some functions $\Gamma_{ij}^k$ $$ \frac{\partial e'_i}{\partial y^j}=\Gamma_{ij}^k e_k'.$$ Upon taking inner product of this equation with $e_l'$, one sees that these coefficients are given by $$\Gamma_{ij}^k=g^{kl}e_l'\cdot \frac{\partial e'_i}{\partial y^j}.$$

Now, in index notation, the covariant derivative of $X^i$ is given by the

$$\nabla_jX^i=\frac{\partial X^i}{\partial y^j}+\Gamma_{jk}^iX^k.$$ This is of the form $\frac {{\mathcal D}f(x)}{dx}=\frac {df(x)} {dx} +\delta f(x)$, but $f$ must be a vector field (or higher rank tensor), otherwise the covariant and ordinary derivatives concide.

$${\mathcal D}_{X} V=\lim_{\Delta x \to o}\frac {\Gamma(\gamma)^o_{\Delta x}V_{(\gamma)\Delta x}-V_{(\gamma)o}}{\Delta x},$$ where the ${\mathcal D}_{X}=\nabla_{X}$