You don't want to prove that $\Delta \cap \Delta'$ is a simplex for each $\Delta \in K$ and $\Delta' \in L$, because it's false.
For a counterexample, take a regular hexagon in $\mathbb R^2$ with vertices $A,B,C,D,E,F$, let $K=\Delta$ be the 2-simplex with vertices $A,C,E$, and let $L=\Delta'$ be the 2-simplex with vertices $B,D,F$, and so $\Delta \cap \Delta'$ is a smaller regular hexagon. Of course that smaller hexagon is the realization of a simplicial complex, in fact any regular hexagon is the realization of a simplicial complex with one extra vertex in the center, and six 2-simplices altogether.
You should be able to find a simplicial complex structure on $|K| \cap |L|$ by carefully keeping this subdivision concept in mind. It'll be messy, if you want to do it with full rigor. But the key idea should be that the intersection of any simplex $\Delta$ of $K$ and any simplex $\Delta'$ of $L$ can be subdivided into a simplicial complex. Of course you have to do these subdivisions over all choices of $\Delta$ and $\Delta'$, to make sure that they all fit together into a simplicial complex structure on $|K| \cap |L|$.
Using the assumption that $K$ and $L$ are finite simplicial complexes, I think the best approach to writing down a rigorous proof is probably as a double induction. For the primary induction, you start with $L = \emptyset$ and then add one simplex of $L$ at a time, writing $L$ as the union of subcomplexes
$$L_1 \subset L_2 \subset \cdots \subset L_I
$$
where each $L_i$ is obtained from $L_{i-1}$ by adding a single simplex $\tau_i$. Assuming that $K$ and $L_{i-1}$ have been subdivided so that their intersection is a subcomplex of each, then you add $\tau_i$. Now you do a secondary induction on the dimension of the skeleta of $K$, writing $K$ as the union of its skeleta $K^{(0)} \subset K^{(1)} \subset \cdots \subset K^{(d)}=K$. You first subdivide $\tau_i$ at each point where its interior intersects $K^{(0)}$, after which $K^{(0)} \cap L_i$ will be a subcomplex of each of $K^{(0)}$ and $L_i$. In the induction step, assuming that $K^{(j)} \cap L_i$ is a subcomplex of each of $K^{(j)}$ and $L_i$, you then further subdivide $\tau_i$ so that $K^{(j+1)} \cap L_i$ is a subcomplex of each of $K^{(j+1)}$ and $L_i$.
So I'm not aware of "simplicial manifold" as a thing. On the other hand the category of "PL manifolds" is a thing, and its definition is slightly more intricate than what you are saying, namely a PL version of 2: A simpicial $n$-complex is a PL-manifold if the link of every vertex is a PL-manifold of dimension $n-1$ that is PL-homeomorphic to the standard PL manifold structure on the sphere of dimension $n-1$ (e.g. the boundary of the $n$-simplex). And then this definition does indeed imply the correponding version of 1. Here's a quick proof, by induction on $n$.
Consider a PL manifold $M$ and a $k$-simplex $\sigma \subset M$, with $k \ge 1$. Pick a $0$-simplex $v \in \sigma$ and therefore $S = \text{Link}_M(v)$ is an $n-1$-sphere with the standard PL structure. The intersection $\tau = \sigma \cap S$ is a $k-1$ simplex in $S$. The key observation is that $\text{Link}_M(\sigma)$ is simplicially isomorphic to $\text{Link}_S(\tau)$, and so it follows by induction on $n$ that this link is a standard PL sphere of dimension $n-k-1$.
But in general, 2 does not imply 1. A counterexample is given by the double suspension theorem of Cannon and Edwards, which produces a simplicial structure on $S^5$ that satisfies 2, and a 1-simplex in that simplicial structure whose link is a 3-dimensional manifold that is not even homeomorphic to $S^3$.
Best Answer
They appear not to be equivalent. Take two tetrahedra (full complexes on four vertices) and join at a point. This gives us a simplicial complex on $7$ vertices whose geometric realisation is contractible.
By your first definition, this simplicial complex is $2$-connected. By your second, it appears not to be. It's not even $1$-connected.
A more combinatorial way of describing the first definition might be to say something like:
Two simplices $\sigma$ and $\tau$ are $k$-connected if there is a sequence of simplices $\sigma,...,\tau$ such that any two consecutive simplices are joined by a simplex.
However, I wouldn't go just altering your definitions without very strong motivation - whilst I know little about the topic of the second paper, there is sure to be good reasoning behind their definition.