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This is not true in the non-semisimple case. Here is an example from finite group representation theory.

Let $G$ be $A_5\cong \operatorname{SL}(2,4)$, and let $k$ be an algebraically closed field of characteristic two. Then there are three simple modules in the principal block of $kG$, which we denote $k$, $M$ and $N$, where $k$ is the trivial module and $M$ and $N$ are self dual two dimensional modules related by the Frobenius map. The structure of the projective cover of the trivial module is as follows:

$$ \begin{matrix} &k\\M&&N\\k&&k\\N&&M\\&k \end{matrix} $$

This module is self dual, and has the following self dual submodules:

$$\begin{matrix} k\\N\\k \end{matrix}\qquad\text{and}\qquad \begin{matrix} &k\\N&&M\\&k \end{matrix}$$

Their intersection is the non self dual module

$$\begin{matrix}N\\k\end{matrix}$$

The same group can be used to give a counterexample in the split case. Consider the module

$$\begin{matrix}k\\N\\k\end{matrix}\oplus \begin{matrix}k\\N\\k\end{matrix}$$

One submodule is the left hand summand. The other is the summand generated by an generator for the left summand plus a generator for the socle of the right summand. The intersection of these submodules is again

$$\begin{matrix}N\\k\end{matrix}$$

In fact, you can even do this second example for the symmetric group $S_3$ in characteristic three, where $N$ is interpreted as the sign representation.

Over $\mathbb{C}$, there's a supersymmetric example which goes as follows. Let $E$ be an exterior algebra on a two dimensional vector space $N$, and let $SL(2,3)$ act irreducibly on $N$ as the binary tetrahedral group. Then $SL(2,3)$ acts on $E$, and we can form the semidirect product $G=E\rtimes SL(2,3)$ as a finite supergroup scheme. The projective cover of the trivial $\mathbb{C}G$-module $\mathbb{C}$ has structure

$$\begin{matrix}k\\N\\k\end{matrix}$$

because $SL(2,3)$ acts on $N$ with determinant one, and we can then do the same construction as above.

This is not true in the non-semisimple case. Here is an example from finite group representation theory.

Let $G$ be $A_5\cong \operatorname{SL}(2,4)$, and let $k$ be an algebraically closed field of characteristic two. Then there are three simple modules in the principal block of $kG$, which we denote $k$, $M$ and $N$, where $k$ is the trivial module and $M$ and $N$ are self dual two dimensional modules related by the Frobenius map. The structure of the projective cover of the trivial module is as follows:

$$ \begin{matrix} &k\\M&&N\\k&&k\\N&&M\\&k \end{matrix} $$

See the footnote at the bottom of this post for the notation.

This module is self dual, and has the following self dual submodules:

$$\begin{matrix} k\\N\\k \end{matrix}\qquad\text{and}\qquad \begin{matrix} &k\\N&&M\\&k \end{matrix}$$

Their intersection is the non self dual module

$$\begin{matrix}N\\k\end{matrix}$$

The same group can be used to give a counterexample in the split case. Consider the module

$$\begin{matrix}k\\N\\k\end{matrix}\oplus \begin{matrix}k\\N\\k\end{matrix}$$

One submodule is the left hand summand. The other is the summand generated by an generator for the left summand plus a generator for the socle of the right summand. The intersection of these submodules is again

$$\begin{matrix}N\\k\end{matrix}$$

In fact, you can even do this second example for the symmetric group $S_3$ in characteristic three, where $N$ is interpreted as the sign representation.

Over $\mathbb{C}$, there's a supersymmetric example which goes as follows. Let $E$ be an exterior algebra on a two dimensional vector space $N$, and let $SL(2,3)$ act irreducibly on $N$ as the binary tetrahedral group. Then $SL(2,3)$ acts on $E$, and we can form the semidirect product $G=E\rtimes SL(2,3)$ as a finite supergroup scheme. The projective cover of the trivial $\mathbb{C}G$-module $\mathbb{C}$ has structure

$$\begin{matrix}k\\N\\k\end{matrix}$$

because $SL(2,3)$ acts on $N$ with determinant one, and we can then do the same construction as above.

Footnote: Notation for modules. The OP has asked me to explain the notation for modules.

When I write

$$\begin{matrix}A\\B\\C\end{matrix}$$

what I mean is a uniserial module with composition factors $A$, $B$, $C$. So there is a unique submodule, isomorphic to $C$; modulo this there is a unique submodule, isomorphic to $B$; and modulo this, it's the simple module $A$.

The structure of the projective cover of the trivial module for $A_5$ given above is that there is a unique top composition factor, $k$, and a unique bottom composition factor, also $k$, and the radical modulo the socle is a direct sum of two uniserial modules. Similarly, the second self dual submodule of this has a unique top and unique bottom composition factors isomorphic to $k$, and the radical modulo the socle is $M\oplus N$.

There is a theory of diagrams for modules, developed in my paper with Carlson, "Diagrammatic methods for modular representation theory and cohomology", of which this is part. Beware though that not all modules have a nice diagram of this form.

This is not true in the non-semisimple case. Here is an example from finite group representation theory.

Let $G$ be $A_5\cong \operatorname{SL}(2,4)$, and let $k$ be an algebraically closed field of characteristic two. Then there are three simple modules in the principal block of $kG$, which we denote $k$, $M$ and $N$, where $k$ is the trivial module and $M$ and $N$ are self dual two dimensional modules related by the Frobenius map. The structure of the projective cover of the trivial module is as follows:

$$ \begin{matrix} &k\\M&&N\\k&&k\\N&&M\\&k \end{matrix} $$

This module is self dual, and has the following self dual submodules:

$$\begin{matrix} k\\N\\k \end{matrix}\qquad\text{and}\qquad \begin{matrix} &k\\N&&M\\&k \end{matrix}$$

Their intersection is the non self dual module

$$\begin{matrix}N\\k\end{matrix}$$

The same group can be used to give a counterexample in the split case. Consider the module

$$\begin{matrix}k\\N\\k\end{matrix}\oplus \begin{matrix}k\\N\\k\end{matrix}$$

One submodule is the left hand summand. The other is the summand generated by an generator for the left summand plus a generator for the socle of the right summand. The intersection of these submodules is again

$$\begin{matrix}N\\k\end{matrix}$$

In fact, you can even do this second example for the symmetric group $S_3$ in characteristic three, where $N$ is interpreted as the sign representation.

Over $\mathbb{C}$, there's a supersymmetric example which goes as follows. Let $E$ be an exterior algebra on a two dimensional vector space $N$, and let $SL(2,3)$ act irreducibly on $N$ as the binary octahedral group. Then $SL(2,3)$ acts on $E$, and we can form the semidirect product $G=E\rtimes SL(2,3)$ as a finite supergroup scheme. The projective cover of the trivial $\mathbb{C}G$-module $\mathbb{C}$ has structure

$$\begin{matrix}k\\N\\k\end{matrix}$$

because $SL(2,3)$ acts with determinant one (its restriction to $E$ is the regular representation), and we can then do the same construction as above.

This is not true in the non-semisimple case. Here is an example from finite group representation theory.

Let $G$ be $A_5\cong \operatorname{SL}(2,4)$, and let $k$ be an algebraically closed field of characteristic two. Then there are three simple modules in the principal block of $kG$, which we denote $k$, $M$ and $N$, where $k$ is the trivial module and $M$ and $N$ are self dual two dimensional modules related by the Frobenius map. The structure of the projective cover of the trivial module is as follows:

$$ \begin{matrix} &k\\M&&N\\k&&k\\N&&M\\&k \end{matrix} $$

This module is self dual, and has the following self dual submodules:

$$\begin{matrix} k\\N\\k \end{matrix}\qquad\text{and}\qquad \begin{matrix} &k\\N&&M\\&k \end{matrix}$$

Their intersection is the non self dual module

$$\begin{matrix}N\\k\end{matrix}$$

The same group can be used to give a counterexample in the split case. Consider the module

$$\begin{matrix}k\\N\\k\end{matrix}\oplus \begin{matrix}k\\N\\k\end{matrix}$$

One submodule is the left hand summand. The other is the summand generated by an generator for the left summand plus a generator for the socle of the right summand. The intersection of these submodules is again

$$\begin{matrix}N\\k\end{matrix}$$

In fact, you can even do this second example for the symmetric group $S_3$ in characteristic three, where $N$ is interpreted as the sign representation.

Over $\mathbb{C}$, there's a supersymmetric example which goes as follows. Let $E$ be an exterior algebra on a two dimensional vector space $N$, and let $SL(2,3)$ act irreducibly on $N$ as the binary tetrahedral group. Then $SL(2,3)$ acts on $E$, and we can form the semidirect product $G=E\rtimes SL(2,3)$ as a finite supergroup scheme. The projective cover of the trivial $\mathbb{C}G$-module $\mathbb{C}$ has structure

$$\begin{matrix}k\\N\\k\end{matrix}$$

because $SL(2,3)$ acts on $N$ with determinant one, and we can then do the same construction as above.

This is not true in the non-semisimple case. Here is an example from finite group representation theory.

Let $G$ be $A_5\cong \operatorname{SL}(2,4)$, and let $k$ be an algebraically closed field of characteristic two. Then there are three simple modules in the principal block of $kG$, which we denote $k$, $M$ and $N$, where $k$ is the trivial module and $M$ and $N$ are self dual two dimensional modules related by the Frobenius map. The structure of the projective cover of the trivial module is as follows:

$$ \begin{matrix} &k\\M&&N\\k&&k\\N&&M\\&k \end{matrix} $$

This module is self dual, and has the following self dual submodules:

$$\begin{matrix} k\\N\\k \end{matrix}\qquad\text{and}\qquad \begin{matrix} &k\\N&&M\\&k \end{matrix}$$

Their intersection is the non self dual module

$$\begin{matrix}N\\k\end{matrix}$$

The same group can be used to give a counterexample in the split case. Consider the module

$$\begin{matrix}k\\N\\k\end{matrix}\oplus \begin{matrix}k\\N\\k\end{matrix}$$

One submodule is the left hand summand. The other is the summand generated by an generator for the left summand plus a generator for the socle of the right summand. The intersection of these submodules is again

$$\begin{matrix}N\\k\end{matrix}$$

In fact, you can even do this second example for the symmetric group $S_3$ in characteristic three, where $N$ is interpreted as the sign representation.

Over $\mathbb{C}$, there's a supersymmetric example which goes as follows. Let $E$ be an exterior algebra on a two dimensional vector space $N$, and let $S_3$ act irreducibly on $N$. Then $S_3$ acts on $E$, and we can form the semidirect product $G=E\rtimes S_3$ as a finite supergroup scheme. The projective cover of the trivial $\mathbb{C}G$-module $\mathbb{C}$ has structure

$$\begin{matrix}k\\N\\k\end{matrix}$$

(its restriction to $E$ is the regular representation), and we can then do the same construction as above.

This is not true in the non-semisimple case. Here is an example from finite group representation theory.

Let $G$ be $A_5\cong \operatorname{SL}(2,4)$, and let $k$ be an algebraically closed field of characteristic two. Then there are three simple modules in the principal block of $kG$, which we denote $k$, $M$ and $N$, where $k$ is the trivial module and $M$ and $N$ are self dual two dimensional modules related by the Frobenius map. The structure of the projective cover of the trivial module is as follows:

$$ \begin{matrix} &k\\M&&N\\k&&k\\N&&M\\&k \end{matrix} $$

This module is self dual, and has the following self dual submodules:

$$\begin{matrix} k\\N\\k \end{matrix}\qquad\text{and}\qquad \begin{matrix} &k\\N&&M\\&k \end{matrix}$$

Their intersection is the non self dual module

$$\begin{matrix}N\\k\end{matrix}$$

The same group can be used to give a counterexample in the split case. Consider the module

$$\begin{matrix}k\\N\\k\end{matrix}\oplus \begin{matrix}k\\N\\k\end{matrix}$$

One submodule is the left hand summand. The other is the summand generated by an generator for the left summand plus a generator for the socle of the right summand. The intersection of these submodules is again

$$\begin{matrix}N\\k\end{matrix}$$

In fact, you can even do this second example for the symmetric group $S_3$ in characteristic three, where $N$ is interpreted as the sign representation.

Over $\mathbb{C}$, there's a supersymmetric example which goes as follows. Let $E$ be an exterior algebra on a two dimensional vector space $N$, and let $SL(2,3)$ act irreducibly on $N$ as the binary octahedral group. Then $SL(2,3)$ acts on $E$, and we can form the semidirect product $G=E\rtimes SL(2,3)$ as a finite supergroup scheme. The projective cover of the trivial $\mathbb{C}G$-module $\mathbb{C}$ has structure

$$\begin{matrix}k\\N\\k\end{matrix}$$

because $SL(2,3)$ acts with determinant one (its restriction to $E$ is the regular representation), and we can then do the same construction as above.