No. Think of a $\mathbb Q$-linear map $f:\mathbb R\to \mathbb Q$.
Yes. For small nonzero $v$ and $w$ write $$ E(v,w)=\frac{f(a+v+w)-f(a+v)-f(a+w)+f(a)-b(v,w)}{|v||w|}. $$
By assumption you have bilinear $b$ such that the limit of $E(v,w)$ as $(v,w)\to (0,0)$ is zero. If you also assume that $f$ is differentiable at $a+v$ for all small enough $v$ then for small nonzero $v$ you can take the limit of $E(v,w)$ as $w\to 0$ and get $$ \frac{df_{a+v}(\frac{w}{|w|})-df_{a}(\frac{w}{|w|})-b(v,\frac{w}{|w|})}{|v|}. $$ The limit of this as $v\to 0$ is zero. Therefore the derivative of $x\mapsto df_x$ at $x=a$ exists and corresponds to $b$.
Note that this implies that you get continuity of $df$ at a point if you have existence of $df$ in a neighborhood of the point and existence of $d^2f$ (in the sense we are exploring) at the point.