I am trying to prove that
$D_{u+v}f(a)=D_{u}f(a)+D_{v}f(a)$, where $a \in \mathbb{R}^n$ for any two vectors $u,v$ in $\mathbb{R}^n$.
I try to use the definition and so forth but I could not prove it. I need to prove this only by using the definition, which is $$D_{v}f(a)=\lim_{t \to 0} \frac{f(a+tv)-f(a)}{t}$$.
You cannot prove this if you just assume that the given function is Gâteaux differentiable. You need full Fréchet differentiability, differentiability for short, if you want to prove your claim. Consider the following example: $$f(x):=\sqrt{x_1^2+x_2^2}\>{\rm sgn}(x_2)\ ,$$ whereby ${\rm sgn}(0):=0$. We have $$D_u f(0)=\lim_{t\to0}{f(t\, u)\over t}=\lim_{t\to0}{|t|\,|u|\>{\rm sgn}(t\,u_2)\over t}=|u|\,{\rm sgn}(u_2)\ .$$ The RHS here is defined for all $u\in{\mathbb R}^2$, but clearly not additive.
This example shows that differentiability is a stronger condition than what you are allowed to assume. If your $f$ is in fact Fréchet differentiable then it is easy to prove your claim: One has $$f(a+ tu)-f(a)=df(a).(t\,u)+o\bigl(|t\, u|\bigr)= t \bigl(df(a).u+o(1)\bigr)$$ and therefore $$\lim_{t\to0}{f(a+ tu)-f(a)\over t}=df(a).u\ ,$$ which is clearly linear in $u$.