Question on "false" application of chain rule Let's consider $K:\mathbb{R}^n\times [0,1]\to\mathbb{R}$ and $f:\mathbb{R}^n\to\mathbb{R}^n$ differentiable with $K(u,t):=\langle f(a+t(u-a)),(u-a)\rangle=\sum\limits_{i=1}^nf_i(a+t(u-a))(u_i-a_i)$, where $a\in\mathbb{R}^n$ .
Now let's take the first partial derivative (with respect to $u_1$) and apply the chain and product rule of differentiation:
\begin{align*}
&D_1K(u,t)=D_1\left(\sum\limits_{i=1}^nf_i(a+t(u-a))(u_i-a_i)\right)\\
&=\sum\limits_{i=1}^nD_1f_i(a+t(u-a))~\underset{=?}{\underbrace{D_1(a+t(u-a))}}~(u_i-a_i)+f_1(a+t(u-a)).\end{align*}
As $(a+t(u-a))$ attains the form of $\begin{pmatrix}a_1+t(u_1-a_1)\\\vdots\\a_n+t(u_n-a_n) \end{pmatrix}$,
I thought that $D_1(a+t(u-a))=D_1\begin{pmatrix}a_1+t(u_1-a_1)\\a_2+t(u_2-a_2)\\ \vdots\\a_n+t(u_n-a_n) \end{pmatrix}=\begin{pmatrix}t\\0\\\vdots\\0 \end{pmatrix}$ but this doesn't match the dimension of the image of $K$. Where is my mistake?

Your application of the chain rule is not entirely correct. We have:
$$D_1\left(f_i(a+t(u-a))\right)=\sum_jD_jf_i(a+t(u-a))D_1(a_j+t(u_j-a_j))=tD_1f_i(a+t(u-a)).$$
So, this then leads to
$$D_1K(u,t)=\sum_iD_1\left(f_i(a+t(u-a))\right)(u_i-a_i)+f_1(a+t(u-a))=t\langle D_1f(a+t(u-a)),(u-a)\rangle+f_1(a+t(u-a)).$$
