Why is the second mixed partial not belong to tangent space $\partial^2_{ij} \not\in T_pM$ So I was just reading on wikipedia about covariant derivatives
And it is mentioned that $\frac{\partial^2}{\partial_i\partial_j}$ is not tangential to $M$, so that means it won't be in the tangent space of $M$. 
But isn't there a canonical isomorphism of $T_0(T_pM) \approx T_pM$?
Or is there a more simple geometric reason that the rate of change of the velocity field must not be in the tangent space?
Every picture (Da Carmo, or Presley) shows that $dv/dt$ is a vector pointing away from the manifold $M$ and tangent space $T_pM$ with it orthogonal decomposition. How do we know $dv/dt$ can't lie in the spaces?
 A: They just mean "not necessarily tangential", i.e., you cannot hope to use that they are tangential in any general argument. But sometimes that $\vec n$ will be $0$.
Here's a simple geometric example. Consider the circle $S^1\subset \mathbb{R}^2$ embedded in the plane, and consider the constant speed path $\gamma:\mathbb{R} \to S^1$ given by $\theta\mapsto (\cos(\theta),\sin(\theta))$. The velocity vector is always tangent to the circle, but the acceleration is always radial, perpendicular to the circle. 
But it's not that these are never in the tangent space. For example, embed $\mathbb{R}$ inside $\mathbb{R}^2$ as the $x$-axis. Any path in this copy of $\mathbb{R}$ necessarily has acceleration which is tangent to the line, because it is straight. But you could also embed $\mathbb{R}$ into $\mathbb{R}^2$ as a parabola by $t\mapsto (t,t^2)$, which will have some acceleration vectors which are not tangent to the parabola.
A: $\frac{\partial^2}{\partial_i\partial_j}$ is not a tangent vector because it does not satisfy the product rule:
$$\frac{\partial^2}{\partial_i\partial_j}(x_ix_j)
=1\neq 0
=(\frac{\partial^2}{\partial_i\partial_j}x_i)\cdot x_j+x_i\cdot (\frac{\partial^2}{\partial_i\partial_j}x_j)$$
