I think it is either assumed that the $f_i$ are defined on the same domain $D$, or that (following a common convention) we set $f_i(x)=+\infty$ if $x \notin \mathrm{Dom}(f_i)$. You can easily check that under this convention, the extended $f_i$ still remain convex and the claim is true.
In general, there is no good way to find the conjugate of a composition. But you can use the following fact:
Fact. Let $M$ be a linear subspace of $\mathbb R^n$, and let $g:M\to\mathbb R\cup\{+\infty\} $ be a convex function. If $\pi_M$ is the orthogonal projection onto $M$, then
$$
(g\circ \pi_M)^*(y) = \begin{cases} g^*(y),\quad &y\in M \\ +\infty,\quad &y\notin M \end{cases} \tag1
$$
Proof. Let $M^\perp$ be the orthogonal complement of $M$. Every vector $x$ decomposes as $x_M+x_{M^\perp}$ where $x_M\in M$ and $x_{M^\perp}\in {M^\perp}$. Therefore,
$$(g\circ \pi_M)^*(y) =\sup_{x_M, \ x_{M^\perp}} \left(\langle y, x_M\rangle+ \langle y, x_{M^\perp}\rangle - f(x_M) \right) \tag2
$$
Observe that $x_{M^\perp}$ appears only in the term $\langle y, x_{M^\perp}\rangle$. If $y\in M$, then $\langle y, x_{M^\perp}\rangle=0$ for all $x_{M^\perp}$ and therefore (2) becomes the formula for $g^*$. If $y\notin M$, then $\sup_{ x_{M^\perp}} \langle y, x_{M^\perp}\rangle =+\infty$. $\Box$
Combining the above fact with your knowledge of one-dimensional conjugate
$$|cx|^*(y)=\begin{cases} 0 , \quad &|y|\le |c| \\ +\infty , \quad &\text{otherwise} \end{cases} \tag3$$
you can conclude with
$$|\langle a,x\rangle|^*(y)=\begin{cases} 0 , \quad &y=ta, \ -1\le t\le 1 \\ +\infty , \quad &\text{otherwise} \end{cases} \tag4$$
Best Answer
Let $(g_i)_{i\in I}$ be a family of convex functions on a convex compact set $\Omega\subseteq \mathbb{R}^d$. We will show that the sup of this family is convex. We will use the standard definition of convexity.
Let $g:=\sup_{i\in I} g_i$.
Take $x,y\in\Omega$ and $t\in[0,1]$.
Fix $i\in I$. Since $g_i$ is convex and bounded above by $g$, we have $$ g_i(tx+(1-t)y)\leq tg_i(x)+(1-t)g_i(y)\leq tg(x)+(1-t)g(y). $$ Since the latter holds for every $i\in I$, we can take the sup and find $$ g(tx+(1-t)y)\leq tg(x)+(1-t)g(y). $$
This holds for every $x,y\in \Omega$ and every $t\in[0,1]$. So $g$ is convex.
Now every affine function $f_i$ is convex, so the result follows from the general case above.
Geometrically? A function is convex iff its epigraph is convex. See here for a definition of the epigraph. It is clear that the epigraph of $\sup g_i$ is the intersection of the epigraphs of all the $g_i$. Now the intersection of convex sets is convex, which yields a more geometric proof of the statement above.