How is the gradient of a function $f$ equal to the Frechet derivative? A mapping $f$ from an open set $S \subset \mathbb{R^n}$ into $\mathbb{R^m}$ is said to be differentiable at $\vec{a} \in S$ if there is an $n\times m$ matrix $L$ such that $\lim_{\vec{h} \to 0}\dfrac{|f(\vec{a}+\vec{h})-f(\vec{a})-L\cdot \vec{h}|}{|\vec{h}|}=0$.
If we look at it component wise, we get $\lim_{\vec{h} \to 0}\dfrac{|f_j(\vec{a}+\vec{h})-f_j(\vec{a})-L^j\cdot \vec{h}|}{|\vec{h}|}=0$ for $j=1,...,m$ where $L^j$ is the row of the matrix $L$. The text book I am reading then states that the components $f_j$ are differentiable at $\vec{x}=\vec{a}$ and that $\nabla f_j(\vec{a})=L^j$. 
I fail to see how $\nabla f_j(\vec{a})=L^j$.

Gradient of function usually means right away that you are restricting to $m=1$ (otherwise we'd be talking about a Jacobian matrix). But even then, it's preferable to call the linearized map $L$ you found as the differential, Frechet or total derivative before associating it to the gradient vector. Take a look at the few paragraphs down in the wiki article.
