That $K$ is a bimodule map means that $\phi(a)K(x) = K(\phi(a)(x))$ for all $a\in A, x\in F$. That is, $K \in \phi(A)' \subseteq \mathcal{L}_B(F)$.

I am here following the ideas of Lance's little Hilbert $C^*$-modules book, Chapter 4. Let $z = \sum_i x_i\otimes y_i \in E\odot F$, so $(z|z) = \sum_{i,j} (y_i|\phi((x_i|x_j))y_j) = (y|\phi_n(X)y)$. Here $X\in M_n(A)$ is the matrix with entries $(x_i|x_j)$, $\phi_n$ is the $n$th amplification (as $\phi$ is a $*$-homomorphism, it is completely positive, and so $\phi_n(X) \in M_n(\mathcal{L}_B(F))$ is positive), and $y$ is column with entries $(y_n)$. Here we have turned $F^n$ (as column vectors) into a Hilbert $C^*$-module over $B$ in the obvious way.

Now consider $z'=(\iota\otimes K)z$ which has $(z'|z') = (y|K^*\phi_n(X)Ky)$ where $K$ acts on $F^n$ pointwise. As $K$ commutes with $\phi_n(X)$, $K^*\phi_n(X)K = K^*K \phi_n(X)$. As $\phi_n(X)$ is positive and $\phi_n(M_n(A)) = M_n(\phi(A))$ is a $C^*$-subalgebra of $M_n(\mathcal{L}_B(F)) = \mathcal{L}_B(F^n)$, there is $T\in M_n(A)^+$ with $\phi_n(T^2) = \phi_n(X)$. Then $K^*K\phi_n(T^2) = \phi_n(T) K^*K \phi_n(T)$. It follows that
\begin{align*} (z'|z') &= (\phi_n(T)y|K^*K \phi_n(T)y) \leq \|K\|^2 (\phi_n(T)y|\phi_n(T)y) \\
&= \|K\|^2 (y|\phi_n(X)y) = \|K\|^2 (z|z). \end{align*}
Hence $(\iota\otimes K)$ is bounded (with norm $\|K\|$).

As always $L\otimes\iota$ is bounded (see Lance's book) we can compose to see that $L\otimes K$ is bounded, as required. (Contrary to the original question, I think it is far from obvious that $L\otimes K$ is bounded).

Once we have established that $L\otimes K$ exists, it follows from just a calculation that it is adjointable, with adjoint $L^*\otimes K^*$. Indeed, $(x\otimes y| (L\otimes K)(x'\otimes y')) = (y|\phi((x|Lx')) Ky') = (y|K \phi((x|Lx')) y')
= (K^*y|\phi((L^*x|x')) y') = ((L^*\otimes K^*)(x\otimes y)|x'\otimes y')$.