Is there a Borel subset of $ \mathbb{R}^{2} $, with finite vertical cross-sections, whose projection onto the first component is non-Borel?

No, no such set exists. This is a special case of the Lusin–Novikov theorem; see e.g. Kechris, Classical Descriptive Set Theory, Theorem 18.10.

In general, let $X,Y$ be standard Borel spaces, and suppose $M \subset X \times Y$ is Borel. (Here we are taking $X=Y=\mathbb{R}$.) For $x \in X$, let $M_x = \{y : (x,y) \in M\}$ be the section of $M$ at $x$. The Lusin–Novikov theorem asserts that if all but at most countably many of the $M_x$ are at most countable (i.e. $|\{x : |M_x| > \aleph_0\}| \le \aleph_0$) , then $M$ has a Borel uniformization: there is a Borel set $M^* \subset M$ such that $M^*$ intersects every nonempty section $M_x$ in exactly one point. In particular, the projection map $\pi : X \times Y \to X$, which is Borel, is injective when restricted to $M^*$; so as a consequence (see Kechris Corollary 15.2) we have that $\pi(M^*) = \pi(M)$ is Borel in $X$.

Any set $M$ satisfying your second condition certainly satisfies the hypothesis of Lusin–Novikov (since $\{x : |M_x| > \aleph_0\} = \emptyset$), so its projection is Borel, and thus it does not satisfy your first condition.

The projection is always Borel provided the set is Borel and each cross-section is at most countable. This is an old theorem of Luzin and/or Novikov, valid for sets in any product of two polish spaces, and the countability can be weakened to sigma-compactness (Novikov and/or Kunugui).

There is a pure recursion theoretical proof of the result. The idea is as follows: By Spector-Gandy theorem, a lightface Borel set $(x,y)$ is an r.e set over $L_{\omega_1^{CK}}[x,y]$. If there are at most countably many $y$'s corresponded to the $x$ for every $x$, then it can be reduced to $L_{\omega_1^{CK}}[x]$. Then the projection to the first section becomes a $\Pi^1_1$ statement.