The rings satisfying your condition (for right modules) are the right pure semisimple rings. There are many equivalent conditions. You can find a lot of information in Section 4.5 of the book

Prest, Mike, Purity, spectra and localisation., Encyclopedia of Mathematics and its Applications 121. Cambridge: Cambridge University Press (ISBN 978-0-521-87308-6/hbk). xxviii, 769 p. (2009). ZBL1205.16002.

or you might find it easier to access the older paper

Prest, Mike, Rings of finite representation type and modules of finite Morley rank, J. Algebra 88, 502-533 (1984). ZBL0538.16025.

As you say, such a ring must be right artinian.

It is known that a ring is both left and right pure semisimple if and only if it has finite representation type (i.e., every module is a direct sum of indecomposable modules, and there are finitely many isomorphism types of indecomposable module), which is a left/right symmetric condition. And there is a longstanding conjecture about a strengthening of this.

Pure Semisimplicity Conjecture: A right pure semisimple ring has finite representation type.

Or equivalently this says that pure semisimplicity should be a left/right symmetric condition. There are many positive results for particular classes of rings.

In Question $2$ you say that you are particularly interested in the hereditary case. This doesn't make the conjecture easier, as Herzog proved that if there is a counterexample then there is a hereditary counterexample. Combining this with a result of Simson, it turns out that to prove the conjecture it would be enough to prove that a right pure semisimple hereditary ring is left artinian.

There is a lot of work on rings of finite representation type, especially hereditary ones. The fundamental result is Gabriel's theorem classifying the finite dimensional algebras over an algebraically closed field with finite representation type as those Morita equivalent to path algebras of quivers whose underlying graph is a disjoint union of simply laced Dynkin diagrams. There are many generalizations; one for general hereditary artinian rings is

Dowbor, P.; Ringel, Claus Michael; Simson, D., Hereditary Artinian rings of finite representation type, Representation theory II, Proc. 2nd int. Conf., Ottawa 1979, Lect. Notes Math. 832, 232-241 (1980). ZBL0455.16013.

The rings satisfying your condition (for right modules) are the right pure semisimple rings. There are many equivalent conditions. You can find a lot of information in Section 4.5 of the book

Prest, Mike, Purity, spectra and localisation., Encyclopedia of Mathematics and its Applications 121. Cambridge: Cambridge University Press (ISBN 978-0-521-87308-6/hbk). xxviii, 769 p. (2009). ZBL1205.16002.

or you might find it easier to access the older paper

Prest, Mike, Rings of finite representation type and modules of finite Morley rank, J. Algebra 88, 502-533 (1984). ZBL0538.16025.

As you say, such a ring must be right artinian.

It is known that a ring is both left and right pure semisimple if and only if it has finite representation type (i.e., every module is a direct sum of indecomposable modules, and there are finitely many isomorphism types of indecomposable module), which is a left/right symmetric condition. And there is a longstanding conjecture about a strengthening of this.

Pure Semisimplicity Conjecture: A right pure semisimple ring has finite representation type.

Or equivalently this says that pure semisimplicity should be a left/right symmetric condition. There are many positive results for particular classes of rings.

In Question $2$ you say that you are particularly interested in the hereditary case. This doesn't make the conjecture easier, as Herzog proved that if there is a counterexample then there is a hereditary counterexample. Combining this with a result of Simson, it turns out that to prove the conjecture it would be enough to prove that a right pure semisimple hereditary ring is left artinian.

There is a lot of work on rings of finite representation type, especially hereditary ones. The fundamental result is Gabriel's theorem classifying the finite dimensional hereditary algebras over an algebraically closed field with finite representation type as those Morita equivalent to path algebras of quivers whose underlying graph is a disjoint union of simply laced Dynkin diagrams. There are many generalizations; one for general hereditary artinian rings is

Dowbor, P.; Ringel, Claus Michael; Simson, D., Hereditary Artinian rings of finite representation type, Representation theory II, Proc. 2nd int. Conf., Ottawa 1979, Lect. Notes Math. 832, 232-241 (1980). ZBL0455.16013.

The rings satisfying your condition (for right modules) are the right pure semisimple rings. There are many equivalent conditions. You can find a lot of information in Section 4.5 of the book

Prest, Mike, Purity, spectra and localisation., Encyclopedia of Mathematics and its Applications 121. Cambridge: Cambridge University Press (ISBN 978-0-521-87308-6/hbk). xxviii, 769 p. (2009). ZBL1205.16002.

or you might find it easier to access the older paper

Prest, Mike, Rings of finite representation type and modules of finite Morley rank, J. Algebra 88, 502-533 (1984). ZBL0538.16025.

As you say, such a ring must be right artinian.

It is known that a ring is both left and right pure semisimple if and only if it has finite representation type (i.e., every module is a direct sum of indecomposable modules, and there are finitely many isomorphism types of indecomposable module), which is a left/right symmetric condition. And there is a longstanding conjecture about a strengthening of this.

Pure Semisimplicity Conjecture: A right pure semisimple ring has finite representation type.

Or equivalently this says that pure semisimplicity should be a left/right symmetric condition. There are many positive results for particular classes of rings.

In Question $2$ you say that you are particularly interested in the hereditary case. This doesn't make the conjecture easier, as Herzog proved that if there is a counterexample then there is a hereditary counterexample. Combining this with a result of Simson, it turns out that to prove the conjecture it would be enough to prove that a right pure semisimple hereditary ring is left artinian.

There is a lot of work on rings of finite representation type, especially hereditary ones. The fundamental result is [Gabriel's theorem][1] classifying the finite dimensional algebras over an algebraically closed field with finite representation type as those Morita equivalent to path algebras of quivers whose underlying graph is a disjoint union of simply laced Dynkin diagrams. There are many generalizations; one for general hereditary artinian rings is

Dowbor, P.; Ringel, Claus Michael; Simson, D., Hereditary Artinian rings of finite representation type, Representation theory II, Proc. 2nd int. Conf., Ottawa 1979, Lect. Notes Math. 832, 232-241 (1980). ZBL0455.16013. [1]: https://en.wikipedia.org/wiki/Gabriel%27s_theorem

The rings satisfying your condition (for right modules) are the right pure semisimple rings. There are many equivalent conditions. You can find a lot of information in Section 4.5 of the book

Prest, Mike, Purity, spectra and localisation., Encyclopedia of Mathematics and its Applications 121. Cambridge: Cambridge University Press (ISBN 978-0-521-87308-6/hbk). xxviii, 769 p. (2009). ZBL1205.16002.

or you might find it easier to access the older paper

Prest, Mike, Rings of finite representation type and modules of finite Morley rank, J. Algebra 88, 502-533 (1984). ZBL0538.16025.

As you say, such a ring must be right artinian.

It is known that a ring is both left and right pure semisimple if and only if it has finite representation type (i.e., every module is a direct sum of indecomposable modules, and there are finitely many isomorphism types of indecomposable module), which is a left/right symmetric condition. And there is a longstanding conjecture about a strengthening of this.

Pure Semisimplicity Conjecture: A right pure semisimple ring has finite representation type.

Or equivalently this says that pure semisimplicity should be a left/right symmetric condition. There are many positive results for particular classes of rings.

In Question $2$ you say that you are particularly interested in the hereditary case. This doesn't make the conjecture easier, as Herzog proved that if there is a counterexample then there is a hereditary counterexample. Combining this with a result of Simson, it turns out that to prove the conjecture it would be enough to prove that a right pure semisimple hereditary ring is left artinian.

There is a lot of work on rings of finite representation type, especially hereditary ones. The fundamental result is Gabriel's theorem classifying the finite dimensional algebras over an algebraically closed field with finite representation type as those Morita equivalent to path algebras of quivers whose underlying graph is a disjoint union of simply laced Dynkin diagrams. There are many generalizations; one for general hereditary artinian rings is

Dowbor, P.; Ringel, Claus Michael; Simson, D., Hereditary Artinian rings of finite representation type, Representation theory II, Proc. 2nd int. Conf., Ottawa 1979, Lect. Notes Math. 832, 232-241 (1980). ZBL0455.16013.