If $f$ is an affine map and $f(0) = 0$, then $f$ is a linear map. If $f$ is affine and $f(0) = 0$, then $f$ is a linear map.
  Define $f(x) = T(x) - T(0)$, where $T$ is an affine map from $R^n$ to $R^k$.


I see that I would have to show that $f(ax) = af(x)$, which means $T(ax) - T(0) = a[T(x)-T(0)]$, but how would I show this equality?

I see that $f(x + y) = f(x) + f(y)$ iff $T(0) = 0$, but I don't see how I would show $f(ax) = af(x)$ since $aT(x) \neq T(ax).$

The definition of an affine map:


  Let $\{p_0, .., p_n\} \subseteq R^n$ be affine independent and let $A$ denote the affine set is spans.  An affine map $T: A \rightarrow R^k$ (for some $k \gt 1)$ is a function satisfying $T(\sum t_ip_i) = \sum t_i T(p_i)$ whenever $\sum t_i = 1$.  The restriction to $[p_0, ..., p_n]$, which is the convex set spanned by $\{p_0, ..., p_n\}$, is also an affine map.
Your thoughts are conspicuously missing the key step of writing down what it means for $T$ to be an affine map.
What do you do, is only reading the first part of the chapter four of the book foundation of optimization by Osman Güler. 
Definition 4.1 A nonempty subset A of a vector space E is called an aﬃne set if for points x and y in A, the line passing through x and y is contained
in A; in other words,
x, y ∈ A ⇒ ℓ := {x + t(y − x) = (1 − t)x + ty : t ∈ R} ⊆ A.
Let A and B be aﬃne sets in vector spaces E and F , respectively. A map
F : A → B is called an aﬃne map if
F ((1 − t)x + ty) = (1 − t)F (x) + tF(y) for all t ∈ R.
Theorem 4.4. If A ⊆ E is an aﬃne subset of E, and a ∈ A is an arbitrary
point, then L := A − a = {y − a : y ∈ A} is a linear subspace of E, which is independent of a ∈ A;consequently,
A = a + L, and L = A − A = {y − z : y, z ∈ A}. 
Let F : A → B be a map between two aﬃne sets A and B. It is easy to
verify that F is an aﬃne map if and only if its graph
gr(F ) := {(x, F (x)) : x ∈ A} ⊆ A × B is an aﬃne set.
Since the gr(F) is an aﬃne set in A × B
Using Theorem 4.4. (0,0) = (a -a,F(a)- F(a) ) = ((a,F(a)) - (a,F(a)) is consequently a linear subspace of L := A × B - A × B providing F(x) is a linear map.