The Borel $\sigma$-algebras generated by these two topologies seem to be equal.

The idea of the proof is as follows. Let $\mathcal M_+$ be the subspace of $\mathcal M$ consisting of measures. It is known that the weak topology on $\mathcal M_+$ is metrizable and the space $\mathcal M_+$ is Polish. Consider the subspace $$\mathcal P=\{(\lambda,\mu)\in\mathcal M_+\times\mathcal M_+:\lambda\perp\mu\}.$$ The symbol $\lambda\perp\mu$ means that there are disjoint $\sigma$-compact subsets $A,B\subseteq\mathbb R^d$ such that $\lambda(A)=\lambda(\mathbb R^d)$, $\mu(B)=\mu(\mathbb R^d)$ and $\lambda(B)=\mu(A)=0$. It can be shown that the set $\mathcal P$ is Borel (of type $F_{\sigma\delta}$) in $\mathcal M_+\times\mathcal M_+$.

Now consider the map $$r:\mathcal P\to\mathcal M,\quad r:(\lambda,\mu)\mapsto\lambda-\mu$$and observe that it is continuous and bijective (as each sign-measure uniquely decomposes into its positive and negative parts).

For any open set $U\subseteq \mathcal M$ the preimage $r^{-1}[U]$ is an open set in $\mathcal P$. By the classical Lusin-Souslin Theorem (15.1 in Kechris' book), the image of any Borel subset of $\mathcal P$ under the injective continuous map $r$ is Borel with respect to the metrizable separable topology $\tau_d$. In particular, the set $U=r[r^{-1}[U]]$ is Borel in the topology $\tau_d$. This implies that the topologies $\tau$ and $\tau_d$ on $\mathcal M$ generate the same Borel $\sigma$-algebras.

The Borel $\sigma$-algebras generated by these two topologies seem to be equal.

The idea of the proof is as follows. Let $\mathcal M_+$ be the subspace of $\mathcal M$ consisting of measures. It is known that the weak topology on $\mathcal M_+$ is metrizable and the space $\mathcal M_+$ is Polish. Consider the subspace $$\mathcal P=\{(\lambda,\mu)\in\mathcal M_+\times\mathcal M_+:\lambda\perp\mu\}.$$ The symbol $\lambda\perp\mu$ means that there are disjoint $\sigma$-compact subsets $A,B\subseteq\mathbb R^d$ such that $\lambda(A)=\lambda(\mathbb R^d)$, $\mu(B)=\mu(\mathbb R^d)$ and $\lambda(B)=\mu(A)=0$. It can be shown that the set $\mathcal P$ is Borel (of type $F_{\sigma\delta}$) in $\mathcal M_+\times\mathcal M_+$.

Now consider the map $$r:\mathcal P\to\mathcal M,\quad r:(\lambda,\mu)\mapsto\lambda-\mu$$and observe that it is continuous and bijective (as each sign-measure uniquely decomposes into its positive and negative parts). Since $\sup_{k\in\mathbb N}\|f_k\|<\infty$, the map $r$ also is also continuous with respect to the topology $\tau_d$ on $\mathcal M$.

Since the Tychonoff space $\mathcal M$ is a continuous image of the metrizable separable space $\mathcal P$, it has countable network of the topology and hence admits a continuous injective map $\psi:\mathcal M\to \mathbb R^\omega$ to the Polish space $\mathbb R^\omega$.

For any $\tau_d$-open set $U\subseteq \mathcal M$ the preimage $r^{-1}[U]$ is an open set in $\mathcal P$. By the classical Lusin-Souslin Theorem (15.1 in Kechris' book), the image of any Borel subset of $\mathcal P$ under the injective continuous map $\psi\circ r$ is Borel in the Polish space $\mathbb R^\omega$. In particular, the set $V=\psi\circ r[r^{-1}[U]]$ is Borel in $\mathbb R^\omega$ and hence the set $U=\psi^{-1}[V]$ is Borel in $\mathcal M$. This implies that the Borel $\sigma$-algebra $\sigma(\tau_d)$ generated by the topology $\tau_d$ is contained in the Borel $\sigma$-algebra $\sigma(\tau)$ generated by the topology $\tau$. On the other hand, the inclusion $\sigma(\tau)\subseteq \sigma(\tau_d)$ follows from the metrizability of the topology $\tau_d$ and the sequential continuity of the indentity map $(\mathcal M,\tau_d)\to\mathcal M$.

The Borel $\sigma$-algebras generated by these two topologies seem to be equal.

The idea of the proof is as follows. Let $\mathcal M_+$ be the subspace of $\mathcal M$ consisting of measures. It is known that the weak topology on $\mathcal M_+$ metrizable and the space $\mathcal M_+$ is Polish. Consider the subspace $$\mathcal P=\{(\lambda,\mu)\in\mathcal M_+\times\mathcal M_+:\lambda\perp\mu\}.$$ The symbol $\lambda\perp\mu$ means that there are disjoint $\sigma$-compact subsets $A,B\subseteq\mathbb R^d$ such that $\lambda(A)=\lambda(\mathbb R^d)$, $\mu(B)=\mu(\mathbb R^d)$ and $\lambda(B)=\mu(A)=0$. It can be shown that the set $\mathcal P$ is Borel (of type $F_{\sigma\delta}$) in $\mathcal M_+\times\mathcal M_+$.

Now consider the map $$r:\mathcal P\to\mathcal M,\quad r:(\lambda,\mu)\mapsto\lambda-\mu$$and observe that it is continuous and bijetive (as each sign-measure uniquely decomposes into its positive and negative parts).

For any open set $U\subseteq \mathcal M$ the preimage $r^{-1}[U]$ is an open set in $\mathcal P$. By the classical Lusin-Souslin Theorem (15.1 in Kechris' book), the image of any Borel subset of $\mathcal P$ under the injective continuous map $r$ is Borel with respect to the metrizable separable topology $\tau_d$. In particular, the set $U=r[r^{-1}[U]]$ is Borel in the topology $\tau_d$. This implies that the topologies $\tau$ and $\tau_d$ on $\mathcal M$ generate the same Borel $\sigma$-algebras.

The Borel $\sigma$-algebras generated by these two topologies seem to be equal.

The idea of the proof is as follows. Let $\mathcal M_+$ be the subspace of $\mathcal M$ consisting of measures. It is known that the weak topology on $\mathcal M_+$ is metrizable and the space $\mathcal M_+$ is Polish. Consider the subspace $$\mathcal P=\{(\lambda,\mu)\in\mathcal M_+\times\mathcal M_+:\lambda\perp\mu\}.$$ The symbol $\lambda\perp\mu$ means that there are disjoint $\sigma$-compact subsets $A,B\subseteq\mathbb R^d$ such that $\lambda(A)=\lambda(\mathbb R^d)$, $\mu(B)=\mu(\mathbb R^d)$ and $\lambda(B)=\mu(A)=0$. It can be shown that the set $\mathcal P$ is Borel (of type $F_{\sigma\delta}$) in $\mathcal M_+\times\mathcal M_+$.

Now consider the map $$r:\mathcal P\to\mathcal M,\quad r:(\lambda,\mu)\mapsto\lambda-\mu$$and observe that it is continuous and bijective (as each sign-measure uniquely decomposes into its positive and negative parts).

For any open set $U\subseteq \mathcal M$ the preimage $r^{-1}[U]$ is an open set in $\mathcal P$. By the classical Lusin-Souslin Theorem (15.1 in Kechris' book), the image of any Borel subset of $\mathcal P$ under the injective continuous map $r$ is Borel with respect to the metrizable separable topology $\tau_d$. In particular, the set $U=r[r^{-1}[U]]$ is Borel in the topology $\tau_d$. This implies that the topologies $\tau$ and $\tau_d$ on $\mathcal M$ generate the same Borel $\sigma$-algebras.