I see no explicit requirement on pages 76-77 that the function be defined on all of $X$, though I agree that it does seem implicit. It seems to me that the obvious far interpretation here is that the statement says that the natural domain of $f+g$ is a measurable subset, and the function is measurable on the domain. Likewise with $fg$, which may not be defined at points where one of the two is $\pm\infty$ and the other is $0$. Also, I did not say anything about points where $f$ or $g$ are not defined. So, what do I mean?

For $f+g$: Let $M=f^{-1}(\infty)$, $N=f^{-1}(-\infty)$, $K=g^{-1}(\infty)$, and $L=g^{-1}(-\infty)$. Each is measurable, since $f$ and $g$ are measurable. Now let $Y = X - ((N\cap K)\cup (M\cap L))$. This is the natural domain of $f+g$, and is measurable. Consider first the restrictions to $Y-(M\cup N\cup K\cup L)$, applying the proof given. Then you work out $(f+g)^{-1}(\infty)$ and $(f+g)^{-1}(-\infty)$ directly in terms of $K$, $L$, $M$, and $N$ (and other sets) to get that $f+g$ is measurable on its natural domain. A similar argument, being careful with the case $f(x)=0$ and $g(x)=0$ this time, will yield the case of $fg$.

As to Question 2, since Halmos is assuming throughout that $f(x)$ and $g(x)$ are finite valued functions, the situation does not arise: the equation involves only finite quantities and all operations are defined. This handles the situation in $X-(K\cup L\cup M\cup N)$. Then you need to consider the situation when you are working in $(M\cup N)\cap g^{-1}(0)$, $(M\cup N)\setminus g^{-1}(0)$, and the remaining two cases.

I see no explicit requirement on pages 76-77 that the function be defined on all of $X$, though I agree that it does seem implicit. It seems to me that the obvious interpretation here is that the statement says that the natural domain of $f+g$ is a measurable subset, and the function is measurable on the domain. Likewise with $fg$, which may not be defined at points where one of the two is $\pm\infty$ and the other is $0$. Also, I did not say anything about points where $f$ or $g$ are not defined. So, what do I mean?

For $f+g$: Let $M=f^{-1}(\infty)$, $N=f^{-1}(-\infty)$, $K=g^{-1}(\infty)$, and $L=g^{-1}(-\infty)$. Each is measurable, since $f$ and $g$ are measurable. Now let $Y = X - ((N\cap K)\cup (M\cap L))$. This is the natural domain of $f+g$, and is measurable. Consider first the restrictions to $Y-(M\cup N\cup K\cup L)$, applying the proof given. Then you work out $(f+g)^{-1}(\infty)$ and $(f+g)^{-1}(-\infty)$ directly in terms of $K$, $L$, $M$, and $N$ (and other sets) to get that $f+g$ is measurable on its natural domain. A similar argument, being careful with the case $f(x)=0$ and $g(x)=0$ this time, will yield the case of $fg$.

As to Question 2, since Halmos is assuming throughout that $f(x)$ and $g(x)$ are finite valued functions, the situation does not arise: the equation involves only finite quantities and all operations are defined. This handles the situation in $X-(K\cup L\cup M\cup N)$. Then you need to consider the situation when you are working in $(M\cup N)\cap g^{-1}(0)$, $(M\cup N)\setminus g^{-1}(0)$, and the remaining two cases.