I found a solution. The theorem is false as stated.
Counterexample. Let $c_1, c_2, c_3$ denote the canonical vectors of $\mathbf{R}^3$ and denote
$$
\begin{array}{rclrcl}
V_1 &= &\langle c_1, c_2 + c_3 \rangle, &W_1 &= &\langle c_2 \rangle, \\
V_2 &= &\langle c_1, c_2 \rangle, &W_2 &= &\langle c_3 \rangle.
\end{array}
$$
Then,
$$
V_1 \cap V_2 = \langle c_1 \rangle, \qquad W_1 + W_2 = \langle c_2, c_3 \rangle.
$$
Notice
$$
\left[\begin{matrix}
x \\ y \\z
\end{matrix}\right] = x c_1 + z (c_2 + c_3) + (y-z) c_2 = x c_1 + y c_2 + z c_3,
$$
therefore
$$
P_1\left[\begin{matrix}
x \\ y \\z
\end{matrix}\right] = x c_1 + z (c_2 + c_3) = \left[\begin{matrix}
x \\ z \\z
\end{matrix}\right],
\qquad
P_2\left[\begin{matrix}
x \\ y \\z
\end{matrix}\right] = x c_1 + y c_2 = \left[\begin{matrix}
x \\ y \\0
\end{matrix}\right]
$$
and
$$
P\left[\begin{matrix}
x \\ y \\z
\end{matrix}\right] = x c_1 = \left[\begin{matrix}
x \\ 0 \\0
\end{matrix}\right].
$$
Then,
$$
P_1P_2\left[\begin{matrix}
x \\ y \\z
\end{matrix}\right] = \left[\begin{matrix}
x \\ 0 \\0
\end{matrix}\right]
=
P\left[\begin{matrix}
x \\ y \\z
\end{matrix}\right],
$$
but
$$
P_2P_1\left[\begin{matrix}
x \\ y \\z
\end{matrix}\right] = \left[\begin{matrix}
x \\ z \\0
\end{matrix}\right].
$$
What is going on? Geometrically, every projector $P_{V \mid W}$ is diagonalisable since it is one times the identity on $V$ and it is zero times the identity on $W.$ A well-known necessary and sufficient condition for two diagonalisable linear functions to commute is that they should be simultaneously diagonalisable. This means that for $P_1$ and $P_2$ to commute, $P_2$ must act as the identy or as zero for every vector in $V_1$ and in $W_1$ and vice versa. The vector spaces $V_i$ and $W_i$ can then be constructed so that the "right cancellations" occur for only $P_1P_2$ to give $P$ but not for $P_2P_1.$ In fact, these considerations gave me a hint to prove a true characterisation.
Theorem. Suppose $\mathbf{R}^d = V_1 \oplus W_1 = V_2 \oplus W_2 =(V_1 \cap V_2) \oplus (W_1 + W_2).$ Let $P_i$ denote the projector onto $V_i$ along $W_i.$ Let $P=P_{V_1 \cap V_2 \mid W_1 + W_2}$ denote the projector onto $V_1 \cap V_2$ along $W_1 + W_2.$ A necessary and sufficient condition for $P = P_1 P_2$ is that $V_2 = (V_2 \cap V_1) \oplus (V_2 \cap W_1).$
Proof. Suppose first $P = P_1 P_2.$ For $x \in V_2$ we have $x = v_1 + w_1$ with $v_1 \in V_1$ and $w_1 \in W_1.$ Since $v_1 = P_1 x = P_1 P_2 x = P x \in V_1 \cap V_2,$ we have $w_1 = x - v_1 \in V_2,$ so $w_1 \in V_2 \cap W_1.$ Therefore, $V_2 = (V_2 \cap V_1) \oplus (V_2 \cap W_1).$ Reciprocally, assume that $V_2$ has the stated decomposition, so $x\in V_2$ can be uniquely written as $x = v + w$ with $v \in V_2 \cap V_1$ and $w \in V_2 \cap W_1.$ This is therefore the unique decomposition of $x \in V_1 \oplus W_1,$ and we see $P_1P_2x = P_1x = v = Px$ (true for all $x \in V_2$). For $x \in W_2,$ we clearly have $P_1P_2 x = P_10 = 0$ and $Px = 0.$ Therefore, $P_1P_2 = P.$ QED
With the previous theorem, it is now obvious that $P = P_1 P_2 = P_2 P_1$ if and only if $V_1 = (V_1 \cap V_2) \oplus (V_1 \cap W_2)$ and $V_2 = (V_2 \cap V_1) \oplus (V_2 \cap W_1).$ In the counterexample above, $V_2 = (V_2 \cap V_1) \oplus (V_2 \cap W_2)$ but $V_1 \neq (V_1 \cap V_2) \oplus (V_1 \cap W_2).$
It suffices to assume that $P_1$ is an orthogonal projector and $P_2$ is any matrix that satisfies $0\preceq P_2\preceq I$. Then $0\preceq P_1P_2P_1\,(\preceq P_1IP_1=P_1^2=P_1)\preceq I$. Therefore $\operatorname{tr}(P_1P_2P_1)\le\operatorname{rank}(P_1P_2P_1)$ (if you don't see this, think about the unitary diagonalisation of $P_1P_2P_1$) and in turn
$$
\operatorname{tr}(P_1P_2)
=\operatorname{tr}(P_1^2P_2)
=\operatorname{tr}(P_1P_2P_1)
\le\operatorname{rank}(P_1P_2P_1)
\le\operatorname{rank}(P_1P_2).
$$
Best Answer
$P_1$ and $P_2$ are projections ($P_i^2=P_i$), so they are determined by the space they project onto. That the squared distance is bounded by the rank (I assume that the distance will get larger if you take $5\times4$ matrices instead of $4\times2$) is connected to the fact that all eigenvalues of projections are $0$ or $1$.
Your projections also satisfy $P_i^t=P_i$, so the matrix $P_1-P_2$ is diagonalisable with respect to some orthonormal basis. The eigenvalues all satisfy $-1\le \lambda_j\le 1$, so the square of the Frobenius norm is bounded by the size of the matrix.