if $A=M$ the assertion is indeed classical: normal states are exactly those ultraweakly continuous. But consider the case where $M=B(H)$ and $\varphi$ is the trace. Then the definition subalgebra $A$ is exactly the trace-class operators. Let $\{p_k\}\subset A$ be a maximal net of pairwise orthogonal projections of trace 1 (i.e. $\{e_{kk}\}$ for any choice of matrix units). Then $p_k\to0$ ultraweakly, but $\mbox{Tr}(p_k)=1$ for all $k$. So the trace is not ultraweakly continuous on the definition algebra, only ultraweakly lower-semicontinuous.

The example on 1) is already explicitly represented, so no.

If you take $B\subset B(H)$ to be a monotone complete C$^*$-subalgebra that is not von Neumann, then you can take $f$ to be the identity map on $B$; it will be obviously normal, but it cannot be ultraweakly continuous because then $B$ would be a von Neumann algebra. (Not sure if this is what you were asking for here)

if $A=M$ the assertion is indeed classical: normal states are exactly those ultraweakly continuous. But consider the case where $M=B(H)$ and $\varphi$ is the trace. Then the definition subalgebra $A$ is exactly the trace-class operators. Let $\{p_k\}\subset A$ be a maximal net of pairwise orthogonal projections of trace 1 (i.e. $\{e_{kk}\}$ for any choice of matrix units). Then $p_k\to0$ ultraweakly, but $\mbox{Tr}(p_k)=1$ for all $k$. So the trace is not ultraweakly continuous on the definition algebra, only ultraweakly lower-semicontinuous.

The example on 1) is already explicitly represented, so no.

Still thinking about it.

if $A=M$ the assertion is indeed classical: normal states are exactly those ultraweakly continuous. But consider the case where $M=B(H)$ and $\varphi$ is the trace. Then the definition subalgebra $A$ is exactly the trace-class operators. Let $\{p_k\}\subset A$ be a maximal net of pairwise orthogonal projections of trace 1 (i.e. $\{e_{kk}\}$ for any choice of matrix units). Then $p_k\to0$ ultraweakly, but $\mbox{Tr}(p_k)=1$ for all $k$. So the trace is not ultraweakly continuous on the definition algebra, only ultraweakly lower-semicontinuous.

The example on 1) is already explicitly represented, so no.

I have to think a little more about this one.

if $A=M$ the assertion is indeed classical: normal states are exactly those ultraweakly continuous. But consider the case where $M=B(H)$ and $\varphi$ is the trace. Then the definition subalgebra $A$ is exactly the trace-class operators. Let $\{p_k\}\subset A$ be a maximal net of pairwise orthogonal projections of trace 1 (i.e. $\{e_{kk}\}$ for any choice of matrix units). Then $p_k\to0$ ultraweakly, but $\mbox{Tr}(p_k)=1$ for all $k$. So the trace is not ultraweakly continuous on the definition algebra, only ultraweakly lower-semicontinuous.

The example on 1) is already explicitly represented, so no.

If you take $B\subset B(H)$ to be a monotone complete C$^*$-subalgebra that is not von Neumann, then you can take $f$ to be the identity map on $B$; it will be obviously normal, but it cannot be ultraweakly continuous because then $B$ would be a von Neumann algebra. (Not sure if this is what you were asking for here)