Following @Kiran's suggestion in the comments, I'll outline why the map $U(1)\to U$ induces an isomorphism between cohomology and K-theory in this setting. At the end I'll also explain a different perspective that might be helpful.

The inclusion maps $U(n)\to U(n+1)$ are $(2n-1)$-connected, so the map $U(1)\to U$ is 1-connected. This means that for any 2-dimensional CW complex X, the induced map Map$(X, U(1))\to$ Map$(X, U)$ is $-1$-connected, and in particular a surjection on $\pi_0$ (I believe this fact about connectivity appears in May's Concise Course; it's proven by induction on skeleta of X). So $[X, U(1)]\to [X, U]$ is surjective. (This is a a borderline case of the result I'm quoting and I would recommend checking carefully that it does work... Note that there's no difference between based and unbased homotopy classes of maps in this setting, because the action of $\pi_1 U(1) = \pi_1 U$ is trivial on $\pi_* U(1)$ and on $\pi_* (U)$, as these are groups.) Now, say $X = M^g$, a closed Riemann surface of genus $g$. Knowing in advance that $[M^g, U(1)] = H^1 (M^g; \mathbb Z) = \mathbb{Z}^2g = K^1 (M^g) = [M^g, U]$, this surjection must be an isomorphism (note that the group structures are induced by the group structures on $U(1)$ and $U$, so the function between homotopy sets is a group homomorphism.)

Another way of thinking about $K^1 (M^g)$ is to consider vector bundles over the suspension $\Sigma M^g$. Since the attaching map of the 2-cell in $M^g$ is a commutator in $\pi_1 (\bigvee_{2g} S^1)$, the attaching map of the 3-cell in $\Sigma M^g$ is a commutator in $\pi_2 (\bigvee_{2g} S^2)$, and hence is nullhomotopic. This means $\Sigma M^g \simeq \bigvee_{2g} S^3$, which gives $[M^g, U] \cong (\pi_3 U)^{2g} \cong \mathbb{Z}^{2g}$. This would probably be more helpful if one could give an explicit homotopy equivalence between $\Sigma M^g$ and $\bigvee_{2g} S^3$. I haven't tried to do that.

Following @Kiran's suggestion in the comments, I'll outline why the map $U(1)\to U$ induces an isomorphism between cohomology and K-theory in this setting. At the end I'll also explain a different perspective that might be helpful.

The inclusion maps $U(n)\to U(n+1)$ are $(2n-1)$-connected, so the map $U(1)\to U$ is 1-connected. This means that for any 2-dimensional CW complex X, the induced map Map$(X, U(1))\to$ Map$(X, U)$ is $-1$-connected, and in particular a surjection on $\pi_0$ (I believe this fact about connectivity appears in May's Concise Course; it's proven by induction on skeleta of X). So $[X, U(1)]\to [X, U]$ is surjective. (This is a a borderline case of the result I'm quoting and I would recommend checking carefully that it does work... Note that there's no difference between based and unbased homotopy classes of maps in this setting, because the action of $\pi_1 U(1) = \pi_1 U$ is trivial on $\pi_* U(1)$ and on $\pi_* (U)$, as these are groups.) Now, say $X = M^g$, a closed Riemann surface of genus $g$. Knowing in advance that $[M^g, U(1)] = H^1 (M^g; \mathbb Z) = \mathbb{Z}^{2g} = K^1 (M^g) = [M^g, U]$, this surjection must be an isomorphism (note that the group structures are induced by the group structures on $U(1)$ and $U$, so the function between homotopy sets is a group homomorphism.)

Another way of thinking about $K^1 (M^g)$ is to consider vector bundles over the suspension $\Sigma M^g$. Since the attaching map of the 2-cell in $M^g$ is a commutator in $\pi_1 (\bigvee_{2g} S^1)$, the attaching map of the 3-cell in $\Sigma M^g$ is a commutator in $\pi_2 (\bigvee_{2g} S^2)$, and hence is nullhomotopic. This means $\Sigma M^g \simeq (\bigvee_{2g} S^2) \vee S^3$, which gives $[\Sigma M^g, BU] \cong \pi_2 (BU)^{2g} \oplus \pi_3 (BU) = \mathbb{Z}^{2g}$ (and by Bott periodicity $[\Sigma M^g, BU] \cong [M^g, \Omega BU] \cong [M^g, U]$). See Tyrone's comment below for a way to make this explicit.

I believe that following @Kiran's suggestion in the comments does give an answer to this question.

I'll outline why the map $U(1)\to U$ induces an isomorphism between cohomology and K-theory in this setting. At the end I'll also explain a different perspective that might be helpful.

The inclusion maps $U(n)\to U(n+1)$ are $(2n-1)$-connected, so the map $U(1)\to U$ is 1-connected. This means that for any 2-dimensional CW complex X, the induced map Map$(X, U(1))\to$ Map$(X, U)$ is $-1$-connected, and in particular a surjection on $\pi_0$ (I believe this fact about connectivity appears in May's Concise Course; it's proven by induction on skeleta of X). So $[X, U(1)]\to [X, U]$ is surjective. (This is a a borderline case of the result I'm quoting and I would recommend checking carefully that it does work... Note that there's no difference between based and unbased homotopy classes of maps in this setting, because the action of $\pi_1 U(1) = \pi_1 U$ is trivial on $\pi_* U(1)$ and on $\pi_* (U)$, as these are groups.) Now, say $X = M^g$, a closed Riemann surface of genus $g$. Knowing in advance that $[M^g, U(1)] = H^1 (M^g; \mathbb Z) = \mathbb{Z}^2g = K^1 (M^g) = [M^g, U]$, this surjection must be an isomorphism (note that the group structures are induced by the group structures on $U(1)$ and $U$, so the function between homotopy sets is a group homomorphism.)

Regarding the question about the Euler pairing: If this just refers to the product in the ring $K^*(M^g)$, I think the product of any two classes in $K^1 (M^g)$ is zero, because the map $H\mathbb Z\to KU$ is a map of ring spectra and hence the induced map from cohomology to $K$-theory respects products, so the product in $K^1 (X)$ is always trivial on the image of $H^1 (X; \mathbb Z)$. (Verifying that the product structure on topological $K$-theory induced from the ring structure on $KU$ agrees with the product structure coming from tensor product of vector bundles is non-trivial, but I think this is discussed in one of May's articles about multiplicative infinite loop space theory from the 70s.)

Another way of thinking about $K^1 (M^g)$ is to consider vector bundles over the suspension $\Sigma M^g$. Since the attaching map of the 2-cell in $M^g$ is a commutator in $\pi_1 (\bigvee_{2g} S^1)$, the attaching map of the 3-cell in $\Sigma M^g$ is a commutator in $\pi_2 (\bigvee_{2g} S^2)$, and hence is nullhomotopic. This means $\Sigma M^g \simeq \bigvee_{2g} S^3$, which gives $[M^g, U] \cong (\pi_3 U)^{2g} \cong \mathbb{Z}^{2g}$. This would probably be more helpful if one could give an explicit homotopy equivalence between $\Sigma M^g$ and $\bigvee_{2g} S^3$. I haven't tried to do that.

Following @Kiran's suggestion in the comments, I'll outline why the map $U(1)\to U$ induces an isomorphism between cohomology and K-theory in this setting. At the end I'll also explain a different perspective that might be helpful.

The inclusion maps $U(n)\to U(n+1)$ are $(2n-1)$-connected, so the map $U(1)\to U$ is 1-connected. This means that for any 2-dimensional CW complex X, the induced map Map$(X, U(1))\to$ Map$(X, U)$ is $-1$-connected, and in particular a surjection on $\pi_0$ (I believe this fact about connectivity appears in May's Concise Course; it's proven by induction on skeleta of X). So $[X, U(1)]\to [X, U]$ is surjective. (This is a a borderline case of the result I'm quoting and I would recommend checking carefully that it does work... Note that there's no difference between based and unbased homotopy classes of maps in this setting, because the action of $\pi_1 U(1) = \pi_1 U$ is trivial on $\pi_* U(1)$ and on $\pi_* (U)$, as these are groups.) Now, say $X = M^g$, a closed Riemann surface of genus $g$. Knowing in advance that $[M^g, U(1)] = H^1 (M^g; \mathbb Z) = \mathbb{Z}^2g = K^1 (M^g) = [M^g, U]$, this surjection must be an isomorphism (note that the group structures are induced by the group structures on $U(1)$ and $U$, so the function between homotopy sets is a group homomorphism.)

Another way of thinking about $K^1 (M^g)$ is to consider vector bundles over the suspension $\Sigma M^g$. Since the attaching map of the 2-cell in $M^g$ is a commutator in $\pi_1 (\bigvee_{2g} S^1)$, the attaching map of the 3-cell in $\Sigma M^g$ is a commutator in $\pi_2 (\bigvee_{2g} S^2)$, and hence is nullhomotopic. This means $\Sigma M^g \simeq \bigvee_{2g} S^3$, which gives $[M^g, U] \cong (\pi_3 U)^{2g} \cong \mathbb{Z}^{2g}$. This would probably be more helpful if one could give an explicit homotopy equivalence between $\Sigma M^g$ and $\bigvee_{2g} S^3$. I haven't tried to do that.