Reconciling different definitions of differentiability in multi-dimensions. We have from this link, the definition of differentiability:
$$\lim_{\mathbf{h}\to \mathbf{0}} \frac{\|\mathbf{f}(\mathbf{x_0}+\mathbf{h}) - \mathbf{f}(\mathbf{x_0}) - \mathbf{J}\mathbf{(h)}\|}{\| \mathbf{h} \|} = 0.$$
Where $J$ is a linear function. This means,
$$\lim_{\mathbf{h}\to \mathbf{0}} \frac{\|\mathbf{f}(\mathbf{x_0}+\mathbf{h}) - \mathbf{f}(\mathbf{x_0}) \|}{\| \mathbf{h} \|} = \lim_{\mathbf{h}\to \mathbf{0}}\frac{\mathbf{J}\mathbf{(h)}}{\|h\|}.$$
Assuming we're coming into the point along the direction $\vec{v}$ (a unit vector), we can say $h=t.\vec{v}$. So we get
$$\lim_{\mathbf{h}\to \mathbf{0}} \frac{\|\mathbf{f}(\mathbf{x_0}+\mathbf{h}) - \mathbf{f}(\mathbf{x_0}) \|}{\| \mathbf{h} \|} = \lim_{\mathbf{t}\to 0}\frac{\mathbf{J}\mathbf{(t\vec{v})}}{t} = \lim_{\mathbf{t}\to 0} \frac{t\mathbf{J}\mathbf{(\vec{v})}}{t} = \mathbf{J}\mathbf{(\vec{v})}.$$
This implies that the limit on the left should be linear in the components of $\vec{v}$. Another way of putting it is that the tangent lines along various directions should lie on a plane. This was the technique @Rene used in the question I asked yesterday.
This is all well and good, but then I saw this other definition of differentiability here. This one says - 
"If, for a function all the partial derivatives of its matrix of partial derivatives exist and are continuous in a neighborhood of the point, it is differentiable". Screenshot - 

Now, I'm finding it hard to reconcile these two definitions. It seems to me the first one is much stricter than the second one. I imagine there might be a function where the derivatives are all continuous, but the tangent lines they correspond to don't lie on a plane. The first condition would seem to imply it's not differentiable while the second one would imply it is. What am I missing here?
 A: Your first limit is the definition. If such a linear function $J$ exists then the function $f$ is differentiable and it can be shown that $J$ is unique.  The existence of the derivative guarantees the existence of all directional (and partial) derivatives
$$D_vf(x) = \lim_{t \to 0}\frac{f(x + t \, v) - f(x)}{t},$$
but the existence of directional derivatives at $x$ is not enough to conclude that $f$ is differentiable by the first definition.
What you are calling the second definition is a theorem that guarantees that the derivative $J$ exists at $x$ -- under fairly strong conditions that the partial derivatives exist in a neighborhood and are continuous at $x$.
An example where the derivative exists at the point $(0,0) \in \mathbb{R}^2$ but the partial derivatives are not continuous at the point is given here.
To prove the theorem for a function $f: \mathbb{R}^n \to \mathbb{R}^m$ where $f(x) = (f_1(x), \ldots, f_n(x))$ it is enough to prove that each component $f_k:\mathbb{R}^n \to \mathbb{R}$ has a derivative.  An example of the proof is given in this answer. 
