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Is it true that the number of submodule of semisimple module is finite? If so, how to explain it?

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No it is not true. It depends on the module.

For example, $M_2(\mathbb R)$ has infinitely many submodules (despite being a module of composition length $2$) while one of its simple submodules has only two submodules.

On the other hand $M_2(F_2)$ has only finitely many submodules.

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    For confirmation, is the $M_2(\mathbb{R})$ regarded as an $M_2(\mathbb{R})$-module? – 2017-01-22
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    I found the submodule and understood now. Thank you. – 2017-01-22
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    @aerile yes, in both cases I meant that the matrix ring is the ring for the module. I think you interpreted it correctly. – 2017-01-22
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No, this is not necessarily true. By definition (see e.g. https://en.wikipedia.org/wiki/Semisimple_module), a semisimple module is the direct sum of simple submodules. But this direct sum may consist of an infinite number of summands: $\bigoplus\limits_{i\in I}M_i$, with an infinite number of $M_i$, in which case by the very construction we have an infinite number of submodules.

(It may be true with additional finiteness conditions, though. But I need to refresh myself on the topic to be able to say more.)