Prove that $ h \cdot (\partial f/\partial g) = df(g;h)$ if $f,g,h$ continuously differentiable

Question

Suppose $f,g,h$ are coontinuously differentiable functions on $\mathbb{R}^n$. Let $f : L_2(\mathbb{R}^n) \to L_2(\mathbb{R}^n)$ be the map $f(g) = f \circ g$. I want to see that the directional derivative $$ df(g;h) = \frac{\partial f}{\partial g} h. $$ My lecturer wrote that $$ df(g;h) = \lim_{t \to 0} \frac{1}{t} \left[ f(g + th) - f(g) \right] = \frac{\partial f}{\partial g} h, $$ because that is the ordinary formula for the directional derivative. I know we have this formula in case that $f$ is a scalar function, and $g,h \in \mathbb{R}^k$, but I am not sure why we get it also in the case that $g,h$ are functions. Can you either show why $\lim_{t \to 0} \frac{1}{t} \left[ f(g + th) - f(g) \right] = \frac{\partial f}{\partial g} h $ or prove it otherwise?