Related to Nate River's answer, I personally prefer to think of Young measures as a single measures $\nu$ on $U \times \mathbb{R}^m$, which have the condition that their projection on the first component (the pushforward along $(x,y) \mapsto x$) is the Lebesgue-measure on $U$. For any reasonable function $f:U\to \mathbb{R}^m$, you can define such a measure "living on" the graph of $f$ in $U \times \mathbb{R}^m$.¹

If you have a bounded sequence of functions $f_k$, using the usual compactness theorems for measures, there is a weakly converging subsequence of the corresponding measures (call them $\nu^{k_j}$) and because of the $L^\infty$ bounds, there is no mass escaping towards infinity, so the projection condition is conserved. The limit $\nu$ then is the Young measure limit (in the above sense) for that sequence.

In particular, now for $G \in C(U\times \mathbb{R}^m)$, we have $$\lim_{k_j \to \infty} \int_U G(x,f_{k_j}(x)) dx = \lim_{k_j \to \infty} \int_{U\times \mathbb{R}^m} G(x,y) d \nu^{k_j} = \int_{U\times \mathbb{R}^m} G(x,y) d \nu $$ just by definition and weak convergence. If you take $G(x,y) = F(y) \phi(x)$ for fixed $F$ and all $\phi$, you recover precisely the theorem in the question.

Now the "classic" Young measure $(\nu_x)_{x\in U}$ then consists of "vertical slices" of this measure, i.e. a disintegration. Since $\{x\} \times \mathbb{R}^m$ always has measure $0$ wrt. to any Young measure defined as above, you can also immediately see, that this is not well defined for any single $x$, but only if you consider enough of them.

If you think graphically about the measures $\nu^{k_j}$ and how they converge you can also see what the measure $\nu$ at $(x_0,y_0)$ (and thus the classic Young measure $\nu_{x_0}$ at $y_0$) represents, namely the limit of how often $f_{k_j}(x)$ for $x$ close to $x_0$ takes values $y$ close to $y_0$.²

There are many other interpretations of Young measures, but most of them are problem specific. If you know what $f_k$ represents, then this gives you a context of what the resulting Young measure limit is supposed to mean. (e.g. a classical example: If $f_k$ represents regularised solutions to a problem, where you force the oscillating stuff to stay on a scale $\frac{1}k$, then you can argue that $\nu_x$ represents the local microstructure, which is too small to resolve on the scale of functions)

There are also a lot of people that offer probabilistic interpretations of Young measures. Personally I am not a fan of those, because there is no randomness involved here. The only reason that $\nu_x$ is a probability measure is because that is the name we give to measures normalized to unit mass.

¹Roughly it is the $n$-dimensional Hausdorff measure restricted to the graph $\{(x,f(x)), x \in U\}$, locally scaled by a factor (something like $\sqrt{1+|\nabla f|^2}^{-1}$) to make the projection work. Alternatively, consider it the pushforward of the Lebesgue-measure on $U$ by $x\mapsto (x,f(x))$

²In particular the $x$ close to $x_0$ part is something that is often missed. Even if $f_{k_j}(x_0)$, converges to $y_0$, the Young measure $\nu_{x_0}$ can be completely different from $\delta_{y_0}$, because of what happens around it for different $x$. (Though only in a sense, as it is not strictly well defined in the first place).

Related to Nate River's answer, I personally prefer to think of Young measures as single measures $\nu$ on $U \times \mathbb{R}^m$, which have the condition that their projection on the first component (the pushforward along $(x,y) \mapsto x$) is the Lebesgue measure on $U$. For any reasonable function $f:U\to \mathbb{R}^m$, you can define such a measure "living on" the graph of $f$ in $U \times \mathbb{R}^m$.¹

If you have a bounded sequence of functions $f_k$, using the usual compactness theorems for measures, there is a weakly converging subsequence of the corresponding measures (call them $\nu^{k_j}$) and because of the $L^\infty$ bounds, there is no mass escaping towards infinity, so the projection condition is conserved. The limit $\nu$ then is the Young measure limit (in the above sense) for that sequence.

In particular, now for $G \in C(U\times \mathbb{R}^m)$, we have $$\lim_{k_j \to \infty} \int_U G(x,f_{k_j}(x)) dx = \lim_{k_j \to \infty} \int_{U\times \mathbb{R}^m} G(x,y) d \nu^{k_j} = \int_{U\times \mathbb{R}^m} G(x,y) d \nu $$ just by definition and weak convergence. If you take $G(x,y) = F(y) \phi(x)$ for fixed $F$ and all $\phi$, you recover precisely the theorem in the question.

Now the "classic" Young measure $(\nu_x)_{x\in U}$ then consists of "vertical slices" of this measure, i.e. a disintegration. Since $\{x\} \times \mathbb{R}^m$ always has measure $0$ wrt. to any Young measure defined as above, you can also immediately see, that this is not well defined for any single $x$, but only if you consider enough of them.

If you think graphically about the measures $\nu^{k_j}$ and how they converge you can also see what the measure $\nu$ at $(x_0,y_0)$ (and thus the classic Young measure $\nu_{x_0}$ at $y_0$) represents, namely the limit of how often $f_{k_j}(x)$ for $x$ close to $x_0$ takes values $y$ close to $y_0$.²

There are many other interpretations of Young measures, but most of them are problem specific. If you know what $f_k$ represents, then this gives you a context of what the resulting Young measure limit is supposed to mean. (e.g. a classical example: If $f_k$ represents regularised solutions to a problem, where you force the oscillating stuff to stay on a scale $\frac{1}k$, then you can argue that $\nu_x$ represents the local microstructure, which is too small to resolve on the scale of functions)

There are also a lot of people that offer probabilistic interpretations of Young measures. Personally I am not a fan of those, because there is no randomness involved here. The only reason that $\nu_x$ is a probability measure is because that is the name we give to measures normalized to unit mass.

¹Roughly it is the $n$-dimensional Hausdorff measure restricted to the graph $\{(x,f(x)), x \in U\}$, locally scaled by a factor (something like $\sqrt{1+|\nabla f|^2}^{-1}$) to make the projection work. Alternatively, consider it the pushforward of the Lebesgue-measure on $U$ by $x\mapsto (x,f(x))$

²In particular the $x$ close to $x_0$ part is something that is often missed. Even if $f_{k_j}(x_0)$, converges to $y_0$, the Young measure $\nu_{x_0}$ can be completely different from $\delta_{y_0}$, because of what happens around it for different $x$. (Though only in a sense, as it is not strictly well defined in the first place).