$\newcommand{\pl}{\partial}$ $\newcommand{\M}{\mathcal{M}}$ $\newcommand{\N}{\mathcal{N}}$
Let $\M,\N$ be smooth compact, connected, oriented manifolds.
Let $f:\M \to \N$ be a smooth immersion. Does there exist $M < \infty$ s.t $|f^{-1}(q)| < \M$ for all $q \in \N$?
(For a proof that $|f^{-1}(q)|$ is finite, see here).
Does anything changes if we assume $\dim \M=\dim \N$?
Edit:
Let me elaborate on the solution of Lee Mosher. The main claim there is that $X_k=\{ q \in \N \, | \, |f^{-1}(q)| \ge k\}$ is closed when $f$ is an immersion. Let's prove this:
Let $q_n \in X_k$, and suppose $q_n \to q \in \N$. Since $|f^{-1}(q_n)| \ge k$, there exist distinct elements $x_n^1,\dots,x_n^k \in \M$ such that $f(x_n^i)=q_n$. Since $\M$ is compact, we can assume W.L.O.G that $x_n^i \to x^i$ for $i=1,\dots,k$. Since $f$ is continuous, $x^i \in f^{-1}(q)$. We claim that all the $x^i$ are distinct. Indeed, assume by contradiction that $x^i=x^j:=x$. Since every immersion is locally an embedding, there exist an open subset $U \subseteq \M$, containing $x$, such that $f|_U$ is injective. Since $x_n^i \to x,x_n^j \to x$, it follows that $x_n^i,x_n^j \in U$ for $n$ large enough, which is a contradiction (to injectivity of $f|_U$ ).
Yes, there is a finite upper bound.
To prove this, let $X_k \subset \mathcal{N}$ be the set of all $q \in \mathcal{N}$ such that $|f^{-1}(q)| \ge k$. Since $f$ is an immersion, the set $X_k$ is a closed subset of $\mathcal{N}$, hence compact.
We have a nested sequence $$X_1 \supset X_2 \supset \cdots $$ If all of the compact sets in this sequence were nonempty, then their intersection would be nonempty, hence there would exist $q \in \mathcal{N}$ such that $|f^{-1}(q)| = \infty$, a contradiction.