In a 3x3 diagram of modules, if the first two rows are split exact, is the third row also split? I have the following commutative diagram of $R$-modules
$$\begin{matrix}
0 & \to & L' & \to & L & \to & L'' & \to & 0\\
&& \downarrow && \downarrow && \downarrow &&\\
0 & \to & M' & \to & M & \to & M'' & \to & 0\\
&& \downarrow && \downarrow && \downarrow &&\\
 &  & N' & \to & N & \to & N''& \to & 0\\
&& \downarrow && \downarrow && \downarrow &&\\
&& 0 && 0 && 0 &&\\
\end{matrix}$$
with exact rows and columns, where the first and second rows are split exact.
Question: Is it also true that the third row is split exact?
I have determined it is true if the splittings are compatible, i.e. the reverse square
$$\begin{matrix}
L & \leftarrow & L''\\
\downarrow && \downarrow\\
M & \leftarrow & M''\\
\end{matrix}$$
is commutative.
However, I'm not sure if that is the case in my situation.
Question 2: If $L'', M''$ are projective, can we choose a splitting of the second row which descends to a splitting of the third row?
Question 3: If the upper left square is Cartesian, and the first two rows split, will the third row split?
I have thought about this for a while, but I can't seem to make the argument work. I would appreciate any help.

Generalizing this answer, every exact sequence $X\xrightarrow{\alpha} Y\xrightarrow{\beta} Z\to 0$ appears as the bottom row of a diagram of the kind you describe, since you can take the diagram
$$\require{AMScd}\begin{CD}
0@>>>0@>>>X@>1>>X@>>>0\\
@.@VVV@VV\begin{pmatrix}1\\\alpha\end{pmatrix}V@VV\alpha V\\
0@>>>X@>\begin{pmatrix}1\\0\end{pmatrix}>>X\oplus Y@>\begin{pmatrix}0&1\end{pmatrix}>>Y@>>>0\\
@.@VV1V@VV\begin{pmatrix}\alpha&-1\end{pmatrix}V@VV-\beta V\\
@.X@>\alpha>>Y@>\beta>>Z@>>>0\\
@.@VVV@VVV@VVV\\
@.0@.0@.0
\end{CD}$$
So no, the third row of your diagram does not have to be split exact: it can be absolutely anything.
Special cases of this also give negative answers to your other questions:
Taking $X\to Y\to Z\to0$ to be a projective presentation of a nonprojective module answers Question 2.
Taking $0\to X\to Y\to Z\to 0$ to be a nonsplit short exact sequence gives an example where the top left square is Cartesian, answering Question 3.
