Directional derivative in the direction of a sum of two vectors.

The counter-example is made of pieces of a cone:

$$z=f(x,y)=\text{sgn}(x)\sqrt{|xy|}.$$

Clearly $f(x,0)=f(0,y)=0$ for all $x$ or $y$, so the directional derivatives along $\mathbf v=(1,0)$ and $\mathbf w=(0,1)$ are both $0$.

But the directional derivative along $\mathbf v+\mathbf w=(1,1)$ is

$$\lim_{t\to0}\frac{f(t,t)-f(0,0)}{t}=\lim_{t\to0}\frac{\text{sgn}(t)\sqrt{t^2}}{t}$$

$$=\lim_{t\to0}\frac{\text{sgn}(t)\,|t|}{t}=\lim_{t\to0}\frac{t}{t}=1.$$

More generally, the directional derivative along $(a,b)$ is

$$\lim_{t\to0}\frac{f(at,bt)-f(0,0)}{t}=\lim_{t\to0}\frac{\text{sgn}(a)\text{sgn}(t)\sqrt{|ab|}\,|t|}{t}$$

$$=\text{sgn}(a)\sqrt{|ab|}.$$

Assuming that $f$ is differentiable at $x=a$ we have

$$f(\mathbf a+t\mathbf v+t\mathbf w)=f(\mathbf a)+\nabla f(\mathbf a)\cdot(t\mathbf v+t\mathbf w)+o(|t\mathbf v+t\mathbf w|)=\\=f(\mathbf a)+\nabla f(\mathbf a)\cdot t\mathbf v+\nabla f(a)\cdot t\mathbf w+o(|t\mathbf v+t\mathbf w|)$$

therefore

$$\frac{ f(\mathbf a+t\mathbf v+t\mathbf w)-f(\mathbf a)}t=\nabla f(\mathbf a)\cdot \mathbf v+\nabla f(\mathbf a)\cdot \mathbf w+\frac{o(|t\mathbf v+t\mathbf w|)}{t}$$

and

$$\lim _{t \rightarrow 0} \frac{f(\mathbf a+t\mathbf v+t\mathbf w)-f(\mathbf{a})}{t}=\nabla f(\mathbf a)\cdot \mathbf v+\nabla f(\mathbf a)\cdot \mathbf w=D_{\mathbf{v}} f(\mathbf{a})+D_{\mathbf{w}} f(\mathbf{a})$$

If $\;f\;$ is differentiable at $\;a\;$ it is pretty easy, since then by linearity of the scalar product (or inner product, to name it with other words) we get

$$D_{v+w}f(a)=\nabla f(a)\cdot(v+w)=\nabla f(a)\cdot v+\nabla f(a)\cdot w=D_vf(a)+D_wf(a)$$

If $\;f\;$ isn't differentiable then it may not be true...but I can't produce a counterexample right now.