Yes, this works, and it follows from Halmos' two projections theorem. This says that if $P$ and $Q$ project onto $M$ and $N$, respectively, then we can decompose the Hilbert space as $$ W = (M\cap N) \oplus (M\cap N^{\perp}) \oplus (M^{\perp}\cap N) \oplus (M^{\perp}\cap N^{\perp}) \oplus (S\oplus T) , $$ and this reduces both $P$ and $Q$, and after a unitary transformation, the action on the last two components (if present) is $$ P = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \quad Q = \begin{pmatrix} 1-H & W \\ W & H \end{pmatrix} , $$ with $0<H<1$ and $W=(H(1-H))^{1/2}$. (This of course generalizes the trivial fact that a projection in $\mathbb C^2$ looks like $\left( \begin{smallmatrix} \cos^2\alpha & \sin\alpha\cos\alpha \\ \sin\alpha\cos\alpha & \sin^2\alpha\end{smallmatrix}\right)$.)

Now if $PQv=QPv$ and $v$ has a component $(v_1,v_2)\in S\oplus T$, then it follows that $Wv_1=Wv_2=0$, so $v_1=v_2=0$. In other words, $V$ must be contained in the sum of the first four summands, and thus the desired modifications can be obtained by simply replacing the part on $S\oplus T$ by zero.

For more than two projections, this argument still works, by considering all pairs of projections.

This works for two projections, and then it follows from Halmos' two projections theorem. I originally thought the same argument would address the general case, but as Dominique pointed out, that's not so clear.

This says that if $P$ and $Q$ project onto $M$ and $N$, respectively, then we can decompose the Hilbert space as $$ W = (M\cap N) \oplus (M\cap N^{\perp}) \oplus (M^{\perp}\cap N) \oplus (M^{\perp}\cap N^{\perp}) \oplus (S\oplus T) , $$ and this reduces both $P$ and $Q$, and after a unitary transformation, the action on the last two components (if present) is $$ P = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \quad Q = \begin{pmatrix} 1-H & W \\ W & H \end{pmatrix} , $$ with $0<H<1$ and $W=(H(1-H))^{1/2}$. (This of course generalizes the trivial fact that a projection in $\mathbb C^2$ looks like $\left( \begin{smallmatrix} \cos^2\alpha & \sin\alpha\cos\alpha \\ \sin\alpha\cos\alpha & \sin^2\alpha\end{smallmatrix}\right)$.)

Now if $PQv=QPv$ and $v$ has a component $(v_1,v_2)\in S\oplus T$, then it follows that $Wv_1=Wv_2=0$, so $v_1=v_2=0$. In other words, $V$ must be contained in the sum of the first four summands, and thus the desired modifications can be obtained by simply replacing the part on $S\oplus T$ by zero.

No. [I think I misinterpreted the question.] Consider for example $$ P_1 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \quad P_2 = \frac{1}{2} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} . $$ Then but up to the sign of the non-diagonal entries, a projection in $W=\mathbb C^2$ is already determined by the $(1,1)$ element.

Yes, this works, and it follows from Halmos' two projections theorem. This says that if $P$ and $Q$ project onto $M$ and $N$, respectively, then we can decompose the Hilbert space as $$ W = (M\cap N) \oplus (M\cap N^{\perp}) \oplus (M^{\perp}\cap N) \oplus (M^{\perp}\cap N^{\perp}) \oplus (S\oplus T) , $$ and this reduces both $P$ and $Q$, and after a unitary transformation, the action on the last two components (if present) is $$ P = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \quad Q = \begin{pmatrix} 1-H & W \\ W & H \end{pmatrix} , $$ with $0<H<1$ and $W=(H(1-H))^{1/2}$. (This of course generalizes the trivial fact that a projection in $\mathbb C^2$ looks like $\left( \begin{smallmatrix} \cos^2\alpha & \sin\alpha\cos\alpha \\ \sin\alpha\cos\alpha & \sin^2\alpha\end{smallmatrix}\right)$.)

Now if $PQv=QPv$ and $v$ has a component $(v_1,v_2)\in S\oplus T$, then it follows that $Wv_1=Wv_2=0$, so $v_1=v_2=0$. In other words, $V$ must be contained in the sum of the first four summands, and thus the desired modifications can be obtained by simply replacing the part on $S\oplus T$ by zero.

For more than two projections, this argument still works, by considering all pairs of projections.