How can we show that $\left\{x\mapsto\prod_{i=1}^kf_i(x_i):f_i\in C_0(E_i)\right\}$ is convergence determining on $\times_{i=1}^ kE_i$? Let $E$ be a complete locally compact separable metric space and $\mu,\mu_n$ be probability measures on $\mathcal B(E)$. We say $\mu_n\to\mu$ weakly if $$\int f\:{\rm d}\mu_n\to\int f\:{\rm d}\mu\tag1$$ for all bounded continuous $f:E\to\mathbb R$ (write $f\in C_b(E)$) and we say $\mu_n\to\mu$ vaguely if $(1)$ for all compactly supported continuous $f:E\to\mathbb R$ (write $f\in C_c(E)$).

By the Portmanteau theorem, $\mu_n\to\mu$ weakly if and only if $\mu_n\to\mu$ vaguely and $\mu(E)=\lim_{n\to\infty}\mu_n(E)$.

Now, we say that $M\subseteq C_b(E)$ is convergence determining if the assertion that $(1)$ holds for all $f\in M$ implies $\mu_n\to\mu$ weakly. By Proposition 4.4 of Chapter 3 in the book of Ethier/Kurtz, $C_c(E)$ is convergence determining. But that means (by our definition) $\mu_n\to\mu$ vaguely.

Question 1: Am I missing something or does that mean that in the Portmanteau theorem we are able to conclude $\mu_n\to\mu$ vaguely and $\mu(E)=\lim_{n\to\infty}\mu_n(E)$ from $\mu_n\to\mu$ weakly, while we only need to know $\mu_n\to\mu$ vaguely in order to conclude $\mu_n\to\mu$ weakly?

Now assume we have complete locally compact separable metric spaces $E_1,\ldots,E_k$. I would like to show that $$\tilde M:=\left\{\prod_{i=1}^k(f_i\circ\pi_i):f_i\in C_0(E_i)\right\},$$ where $\pi_i:\times_{j=1}^dE_j\to E_i$ is the projection onto the $i$th coordinate) is convergence determining (on $\times_{i=1}^kE_i$). Again in Ethier/Kurtz, Proposition 4.6 of Chapter 3, we find the claim that if $M_i\subseteq C_b(E_i)$ is convergence determining (on $E_i$), then $$M:=\left\{\prod_{i=1}^k(f_i\circ\pi_i):f_i\in M_i\cup\color{red}{\left\{1\right\}}\right\}$$ is convergence determining. I would clearly take $M_i=C_0(E_i)$ (which is convergence determining by the former result and since $C_c(E_i)$ is dense in $C_0(E_i)$ by definition). However, Ethier/Kurtz allow $f_i$ to be the constant function $1$ (which is clearly not in $C_0(E_i)$, unless $E_i$ is compact). They state Proposition 4.6 in the more general case of a countable product, but I don't think that this is the reason why we need to allow $f_i\equiv 1$. On the other hand, taking $k=1$ yields a result weaker than what we assume beforehand (we start with $M_1$ being convergence determining and conclude $M_1\cup\left\{1\right\}$ is convergence determining).

Question 2: What am I missing here? Is my $\tilde M$ convergence determining?

 A: Question 1: The point is that in the definition of a set being convergence determining you make the additional assumption that $\mu_n$ and $\mu$ are probability measures and so the thing you call the Portmanteau Theorem (which is true but is not the thing I would call the Portmanteau Theorem) says that $\mu_n \to \mu$ vaguely if and only if $\mu_n \to \mu$ weakly (under the extra assumption). Unfortunately, the space of probability measures isn't closed in the vague topology on the space of, say, finite measures but is closed in the weak topology. To make this concrete, take $E = \mathbb{R}$ and $\mu_n = \delta_n$ to be the point mass at $n$. Then $\mu_n \to 0$ vaguely and not weakly, the problem being that $C_0(E)$ is failing to capture "the escape of mass to infinity" which is exactly what the extra condition in what you call the Portmanteau Theorem fixes.
Question 2: The only thing that we can't do in the same way when we don't allow ourselves the function $1$ in the proof of Proposition 4.6 of Chapter 3 of Ethier and Kurtz is immediately conclude that for $f_i \in M_i$, 
$$\int f_i d\mu_n \to \int f_i d\mu$$
whenever $\mu_n, \mu$ satisfy $(1)$ for all $f \in \tilde{M}$. The extra structure on your equivalent of $M_i$ allows you to get around this. For (notational) convenience, I will demonstrate the idea in the case $k = 2$ and $i = 1$. I imagine this proof can be simplified significantly.
Fix an $\varepsilon > 0$. Also fix strictly increasing sequences $(K_n^i)_{n \geq 1}$ of compact subsets of $E_i$ that cover $E_i$ and satisfy $d(\partial K_n^i, \partial K_{n+1}^i) > 0$ (such compact exhaustions exists since $E_i$ is a locally compact separable metric space, see here). Then since $\mu$ is a probability measure, there is an $N_\varepsilon$ such that 
$$\mu(K_{N_\varepsilon}^1 \times K_{N_\varepsilon}^2) \geq 1 - \frac{\varepsilon}{2}.$$
Now take continuous functions $f_i^{(\varepsilon)}$ valued in $[0,1]$ such that $f_i^{(\varepsilon)} = 1$ on $K_{N_\varepsilon}^i$ and $f_i^{(\varepsilon)} = 0 $ outside of $K_{N_\varepsilon+1}^i $. Then $f_i^{(\varepsilon)} \in C_c(E_i) \subseteq C_0(E_i)$ so
$$\int f_1^{(\varepsilon)} f_2^{(\varepsilon)} d \mu_n \to \int f_1^{(\varepsilon)} f_2^{(\varepsilon)} d \mu \geq 1 - \frac{\varepsilon}{2}$$
so for large enough $n$,
$$\mu_n(K_{N_\varepsilon+1}^1 \times K_{N_\varepsilon+1}^2) \geq \int f_1^{(\varepsilon)} f_2^{(\varepsilon)} d \mu_n \geq 1 - \varepsilon.$$
Now pick continuous, compactly supported functions $g_i^{(\varepsilon)}$ valued in $[0,1]$ such that $g_i^{(\varepsilon)} = 1$ on $K_{N_\varepsilon+1}^i$.
Then, we can bound
\begin{align}
\bigg | \int_{E_1 \times E_2} f_1 d \mu_n - \int_{E_1 \times E_2} f_1 d \mu \bigg| & \leq \bigg | \int f_1 - f_1 g_1^{(\varepsilon)} g_2^{(\varepsilon)} d \mu \bigg| + \bigg | \int   f_1 g_1^{(\varepsilon)} g_2^{(\varepsilon)} d( \mu - \mu_n ) \bigg| \\ &+ \bigg | \int f_1 - f_1 g_1^{(\varepsilon)} g_2^{(\varepsilon)} d \mu_n \bigg|
\end{align}
Now the first and last terms are each bounded by $\|f\|_\infty \varepsilon$ for $n$ sufficiently large, by our choice of $g_i^{(\varepsilon)}$. The middle term goes to $0$ since $f_1 g_1^{(\varepsilon)} \in C_0(E_1)$ and $g_2^{(\varepsilon)} \in C_0(E_2)$ and so for large enough $n$ we have
$$\bigg | \int_{E_1 \times E_2} f_1 d \mu_n - \int_{E_1 \times E_2} f_1 d \mu \bigg| \leq (2 \|f_1\|_\infty + 1) \varepsilon$$
which gives the desired convergence.
