Let $M \neq \lbrace 0 \rbrace$ be a semi-perfect module and $\rho:P \to M$ a projective cover for $M$. I want to prove that $M$ is indecomposable if and only if $P$ is indecomposable.
All right, we understood by a projective cover of $M$ if there is an epimorphism $\rho:P \to M$ with $P$ projective and $\ker( \rho )$ is superfluous on $M$, $\ker(\rho) << P$, that means that for every submodule $L \leq M$ such $M=\ker(\rho)+L$, then $L=M$, i.e, $\ker(\rho)$ is so small..superfluous.
And we understood by $M$ a semi-perfect module if every quotient of $M$ has a projective cover. Also, we say that $M$ is indecomposable if the only diret summands of $M$ are $M$ itself and $\lbrace 0 \rbrace $.
So far I have worked just in only one implication. Let's suppose $M$ is semi-perfect and indecomposable, as there are only two quotients we can take in $M$, these are $\frac{M}{\lbrace 0 \rbrace}=M$ and $\frac{M}{M}=\lbrace 0 \rbrace$ and by hypotheses of $M$ being semi-perfect we have the existence of two projective covers, $\rho_{1}:P_{1} \to M$ and $\rho_{2}:P_{2} \to \lbrace 0 \rbrace$, but I dont know how to relate this two projective cover in order to prove that $P$ in th original projective cover $\rho:P \to M$ of $M$ its indecomposable.
I may well be wrong, but I think only one direction of this is true.
If $M\cong M_{1}\oplus M_{2}$, then since $M_{1}$ and $M_{2}$ are quotients of $M$ they both have projective covers $\phi_{1}:P_{1}\to M_{1}$ and $\phi_{2}:P_{2}\to M_{2}$. Then $\phi_{1}\oplus\phi_{2}:P_{1}\oplus P_{2}\to M$ is also a projective cover by Proposition 5.5.4 of Relative Homological Algebra by Enochs and Jenda. Since projective covers are unique to isomorphism it follows that the projective cover of $M$ is not indecomposable. Hence that direction is true.
I believe I have a counterexample to the other direction. Let $R$ be a local ring and $M$ a finitely generated $R$-module. Then a projective cover of $M$ is given by a map $R^{n}\to M$ where $n$ is the vector space dimension of the $R/\mathfrak{m}$-module $M/\mathfrak{m}M\cong (R/\mathfrak{m})^{n}$ (see Theorem 5.3.3 of RHA). In particular, if $M$ is an indecomposable module then $R^{n}$ will not be indecomposable if $n>1$. But there are local rings with indecomposable f.g. modules $M$ with $n>1$. For example over a complete hypersurface singularity you often find indecomposable maximal Cohen-Macaulay modules with $\mu(M)>1$.