Original definition
A function $f: A \to \mathbb{R}^n$, $A \subseteq \mathbb{R}^m$ is differentiable at a point $\mathbf a \in \mathbb R^m,$ if there is a linear transformation $T$ such that $$ \lim_{\lVert \mathbf h\rVert \to 0} \frac{f(\mathbf a+\mathbf h)-f(\mathbf a)-T_a(\mathbf h)}{\lVert \mathbf h\rVert} = \mathbf 0. $$
My definition
A function $f: A \to \mathbb{R}^n$, $A \subseteq \mathbb{R}^m$ is differentiable at a point $\mathbf a \in \mathbb R^m,$ if there is a linear transformation $T$ such that for any unit vector $\hat u$ $$ \lim_{t \to 0} \frac{f(\mathbf a+t \hat u)-f(\mathbf a)}{t} = T_a(\hat u). $$
Question: is there any difference between original definition and my definition
In the original definition, $\mathbf h$ can approach to $\mathbf 0$ by any trajectory; In my definition it can only be approached from certain direction, so my definition is weaker than the original definition. Substitute $\mathbf h = t \hat u$ to the original definition will get my definition.
If my definition is not true, can somebody provide a counter example ?
One may already noticed that $$ \partial_{\hat u} f (a) = T_a(\hat u) $$ Since $T$ is linear, we can assume (here $\nabla f$ is just a function, it happens to be equal to the gradient if exist): $$ T_a(\hat u) = \hat u \cdot \nabla f(a) $$ So my definition can be written as: $$ \partial_{\hat u} f (a) = \hat u \cdot \nabla f(a) \Leftrightarrow f\text{ is differentiable at }a $$
For a counterexample try $$ f:\mathbb{R}^2 \longrightarrow \mathbb{R} $$ with $$ f(x,y) = \left\{ \begin{array} 11 &\text{if } \:\:x=y^2\:\: \text{ and }\:\:(x,y)\not=(0,0)\\ 0 & \text{otherwise} \end{array} \right. $$
I don't want to prove everything for you. Try to imagine how this function looks like, why it's not differentiable in $0$ (by general definition) and why partial derivatives exist on all directions (following your definition).
EDIT:
Your intuition is very good. The difference between the original definition of differentiability, and the one you proposed is in the fact that the original definition allows $h$ to approach the point $a$ in any trajectory, while you allow it only on linear trajectories.
This counterexample is a function that is $0$ everywhere, except on a small set $A=\{(x,y)\:|\:x=y^2\} \setminus\{(0,0)\}$ where it is equal to $1$. Looking at the function on this curve, it is obviously not continuous in $0$. But it has directional derivatives in all directions in $0$, since for every direction $\hat u$, the point $t\hat u$ is inside $\mathbb{R}^2\setminus A$ for $t$ small enough.