Since it is not always clear what $H^1(X;G)$ means for a non-abelian group, Atiyah might have meant this loosely, perhaps using Cech 1-cocycles to construct flat $G$ bundles, with homologous cycles giving isomorphic bundles. The moduli space of flat bundles is homeomorphic (and real analytically isomorphic) to the space of conjugacy classes of $G$ representations via the holonomy. I suspect this is what Ben is hinting at in his answer above.
But another interpretation in terms of homotopy classes of maps to $BG$ is as follows. If you give $G$ the discrete topology, then $BG$ is a pointed space with a fixed choice of isomorphism $\pi_1(BG)\cong G$. So
$Hom(\pi_1(X),G)$ is in bijective correspondence with the pointed homotopy classes of maps $[X,BG]_0$. The free homotopy classes $[X,BG]$ are then in bijective correspondence with the conjugacy classes $Hom(\pi_1(X),G)/conj$, since the action of $G=\pi_1(BG)$ on the pointed homotopy classes corresponds to conjugation, with quotient the free homotopy classes. So if you define $H^1(X;G)=[X,BG]$ (unbased), you get Atiyah's statement.
In the case of Atiyah's book, though, more important that the notation for the space of conjugacy classes of reps is the fact that at a representation $r:\pi_1(\Sigma)\to G$, the (usual, twisted by the adjoint rep) cohomology $H^1(\Sigma; g_r)$ ($g$ the lie algebra of $G$)is the Zariski tangent space to the variety of conjugacy classes of representations at $r$, i.e. to the moduli space of flat $G$ connections on $\Sigma$. The cup product
$H^1(\Sigma; g_r)\times H^1(\Sigma; g_r)\to R$ determines a symplectic form on this variety, and a holomorphic structure on $\Sigma$ induces a complex structure on this variety, which reflects itself in the Zariski tangent space as an almost complex structure $J:H^1(\Sigma; g_r)\to H^1(\Sigma; g_r)$ which coincides with the Hodge $*$ operator on harmonic forms.
Incidentally, all principal $SU(2)$ bundles over a Riemann surface are topologically trivializable.
Proof of (1):
(a). Suppose $X$ and $Y$ are $G$-spaces, the action of $G$ on $X$ is free, and $X\to X/G$ is
a principal bundle, then the space of $G$-equivariant maps
$$
F(X,Y)^G
$$
is the same thing as the space of sections of the fibration $X\times_G Y \to X/G$.
(b). If $E\to B$ is a Hurewicz fibration, with $B$ connected and the fiber contractible, then the space of sections is again contractible. This can be seen as follows: the adjunction property shows that
$$
\text{sec}(E \to B) \to F(B,E) \to F(B,B)
$$
is a fibration (the displayed fiber is the space of sections---it's the fiber over the basepoint of $F(B,B)$ given by the identity map of $B$---and the other two spaces are function spaces). Since $F$ is contractible the map $E\to B$ is a homotopy equivalence and therefore so is $F(B,E) \to F(B,B)$. Hence $\text{sec}(E\to B)$ is contractible.
(c). It follows from (a) and (b) that the space of sections of $X\times_G Y \to X/G$ is contractible whenever $Y$ is.
(d). In your case, $X = P$ and $Y = EG$. Hence, $F(P,EG)^G$ is contractible.
Proof of (2):
I'll prove something slightly less, which is enough for what
you wish to have:
I claim that
if $X \to X/G$ and $Y \to Y/G$ are principal $G$-bundles then the map
$$
F(X,Y)^G \to F(X/G,Y/G)
$$
is a Hurewicz fibration. This is the same thing as the map
$$
\text{sec}(X\times_G Y\to X/G) \to \text{sec}(X/G\times Y/G\to X/G)
$$
induced by the fibration $X\times_G Y \to X/G \times Y/G$.
This map of section spaces is the map of fibers which is induced by a map of fibrations
over a common base space: the first fibration is $F(X/G, X\times_G Y) \to F(X/G,X/G)$ and the second one is
$F(X/G, X/G\times Y/G) \to F(X/G,X/G)$. Since the map $F(X/G, X\times_G Y) \to F(X/G, X/G\times Y/G)$ is a fibration, it is formal to show that the
map of section spaces is too.
In your situation we take $X = P$ and $Y = EG$. Then we see that the map
$$
F(P,EG)^G \to F(P/G,BG)
$$
is a fibration. The fiber at the point of $F(P/G,BG)$ defined by the classifying map for $P \to P/G$ is then
your gauge group $\cal G$.
Best Answer
This is an edited extract from a book in preparation (Bruner, Catanzaro, May) tentatively titled Characteristic Classes and is therefore overlong for an answer. This is similar to Denis Nardin's answer, but more bundle theoretic; he did refer to an old Memoir of mine, so I thought I'd give an answer. Let $N$ be a closed normal subgroup of a topological group $G$ with quotient group $K$. Let $i\colon N\longrightarrow G$ be the inclusion and $j\colon G\longrightarrow K$ the quotient map. Let $EG \longrightarrow BG$ and $E K \longrightarrow B K$ be universal bundles for $G$ and $K$ and take $EG \longrightarrow EG/ N = B N$ to be the universal bundle for $ N$, so that inclusion of orbits gives $Bi\colon BN \longrightarrow BG$ as a bundle with fiber $K$. There is a map $Ej\colon EG \longrightarrow E K$ such that $(Ej)(yg) = (Ej)(y)j(g)$, either by a functorial construction or general principles. Then passage to orbits gives $Bj\colon BG \longrightarrow BK$. By construction this gives a bundle map [this toy does not seem to tex \xymatrix] from the bundle $Bi$ to the bundle $EK\longrightarrow BK$, so that $BN$ is the pullback along $Bj$ of the universal bundle for $K$. Since $EK$ is contractible, this implies that $BN\longrightarrow BG \longrightarrow BK$ is a fibration sequence. [Further details are standard, but I'll supply if wanted.]