Let $\mathsf{Bimod}$ denote the bicategory whose $0$-cells are rings $A$, a $1$-cell $A \longrightarrow B$ is an $(A,B)$-bimodule and a $2$-cell is a bimodule map. The composition of $1$-cells is given by the tensor product. According to Borceux's Handbook of Categorical Algebra: Volume I, Exercise 7.10.8, this bicategory has left Kan extensions.
So if I take an arbitrary $(A,C)$-bimodule $N$, and an arbitrary $(A,B)$-bimodule $M$, how can I possibly pick a $(C,B)$-bimodule $G$ equipped with a (universal) bimodule map $ M \longrightarrow N \otimes_C G$? Surely the $N$ screws it up somehow, since I have no interaction between $M,N$?
According to Definition 7.1.3 in Borceux's book, the left Kan extension of a bimodule $N : A \to C$ along a bimodule $M : A \to B$ is an initial bimodule $L : B \to C$ equipped with a bimodule map $\alpha : N \to L \circ M$, where $L \circ M : A \to C$ is the bimodule $M \otimes_B L$. In other words, this is a representation of the functor $\hom(N,M \otimes_B -) : {}_B \mathsf{Mod}_C \to \mathsf{Set}$. Any representable functor is continuous, but this functor is usually not continuous when $M$ is "large".
Therefore, left Kan extensions don't exist in $\mathsf{BiMod}$. However, right Kan extensions exist. This is because the functor $\hom(M \otimes_B -,N) : ({}_B \mathsf{Mod}_C)^{\mathrm{op}} \to \mathsf{Set}$ is represented by the $(B,C)$-bimodule $\underline{\hom}_A(M,N)$ of $A$-linear maps $M \to N$ of the underlying $A$-modules with $(b \cdot f \cdot c)(m) := f(m \cdot b) \cdot c$.