On a smooth manifold $M$, a vector field is a smooth map $X : M \to TM$, where $TM$ is the tangent bundle of $M$. If $\chi(M)$ denotes the space of vector fields on $M$, an affine connection $\nabla$ on $M$ is a mapping $\nabla : \chi(M) \times \chi(M) \to \chi(M)$ which (intuitively) creates a new vector field $\nabla_{X}Y$ given two vector fields $X$,$Y$. If I am correct, given $p \in M$, $\nabla_{X}Y(p)$ can be understood as the derivative of $X$ in the direction of $Y(p)$.
1) Consider that $M \subset \mathbb{R}^n$ to make things "easier. Now, a vector field on $M$ is a smooth map $X : M \to \mathbb{R}^n$ (such that, for all $p$, $X(p) \in T_{p}M$). Given another vector field $Y$ and $p \in M$, the derivative of $X$ in the direction of $Y(p)$ could be (naively) $D_{p}X\big( Y(p) \big)$, where $D_{p}X : T_{p}M \to \mathbb{R}^n$ denotes the differential of $X$ at $p$. But this might not define an affine connection since $D_{p}X\big( Y(p) \big)$ is not necessarily a tangent vector, right ?
2) If $\gamma : I \subset \mathbb{R} \to M$ is a smooth curve and $X$ a vector field along $\gamma$, the covariant derivative $\frac{DX}{dt}$ of $X$ (namely $\nabla_{\frac{d\gamma}{dt}}X$) is defined to be the orthogonal projection on the linear space $T_{\gamma(t)}M$. However, this definition holds only because we implicitly assume that $M$ is equipped with the metric induced by $\mathbb{R}^n$, right ? In other words, if I'm correct and if $g$ is a Riemannian metric (which is not the metric induced by $\mathbb{R}^n$) on $M$, this definition of covariant derivative no longer makes sense.