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Let $K, L, E$ be fields with embeddings $K \rightarrow L, K \rightarrow E$. There is in general no canonical field $\Omega$ which contains both $L$ and $E$ as $K$-subfields. Under certain conditions the tensor product $L \otimes_K E$ will be a field, and one can use this.

However, does there always exist some field $\Omega$ which contains $L$ and $E$ as $K$-subfields?

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    Do you really mean "contains both $L$ and $E$ as subfields", or just that $L$ and $E$ *embed* into? – 2017-01-14
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    I mean they should embed in such a way that the compositions $K \rightarrow L \rightarrow \Omega$ and $K \rightarrow E \rightarrow \Omega$ are equal. – 2017-01-14
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    Do you also want that the intersection of the image of $L$ and the image of $E$ in $\Omega$ should be exactly $K$? (That is, *amalgamation*, not just commuting embeddings.) – 2017-01-14
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    That would be nice too, but it's not required. – 2017-01-14
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    Note that full amalgamation is in general impossible, since we could via amalgamation force polynomials of degree $n$ to have as many as $2n$ roots. – 2017-01-14

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Yes, there does. Note that the ring $L\otimes_K E$ is nonzero, since it is the tensor product of two nonzero vector spaces over a field. So it has a maximal ideal $M$. The quotient $\Omega=L\otimes_K E/M$ is then a field, and the canonical inclusions $L\to L\otimes_K E$ and $E\to L\otimes_K E$ give embeddings of $K$-algebras $L\to \Omega$ and $E\to \Omega$.

Alternatively, take $\Omega\supseteq K$ to be algebraically closed and have large transcendence degree over $K$ (at least as large as the transcendence degrees of $L$ and $E$). Picking a transcendence basis $B$ for $L$ over $K$, you can embed $K(B)$ into $\Omega$, and this extends to $L$ since $L$ is algebraic over $K(B)$ and $\Omega$ is algebraically closed. By the same argument there is also an embedding $E\to\Omega$ over $K$.