Q1: In general it is not true that $X^d_t$ is equal to $\sum_{s\le t}\Delta X_s$. In order for this equality to make sense, you need assume something like that the sum of the jumps of $X$ is absolutely convergent (Hypothesis A), although that still does not imply that the identity holds. The term $X^d$ is the purely discontinuous part of the semimartingale $X$, but this terminology can be confusing. A purely discontinuous semimartingale is not in general equal to the sum of its jumps, and need not be discontinuous at all! All (adapted and right-continuous) finite variation processes are "purely discontinuous" semimartingales, regardless of whether they are continuous or not. So, for example, take any continuous finite variation process $X$, such as $X_t=t$. Then, you simply have $X^d_t=X_t-X_0$ whereas $\sum_{s\le t}\Delta X_s=0$. The terminology "purely discontinuous" does make a bit more sense for local martingales, where you can show that purely discontinuous local martingales are, in a sense, orthogonal to the continuous local martingales and are themselves (almost surely) continuous only if they are constant. I've also seen the term "quadratic pure jump semimartingale" to describe purely discontinuous semimartingales (in, e.g., Protter, Stochastic Integration and Differential Equations). This does make some sense, as a semimartingale is purely discontinuous if and only if its quadratic variation is a pure jump process, $[X]_t=\sum_{s\le t}(\Delta X_s)^2$.
Anyway, the short answer to Q1 is that $X^d_t=\sum_{s\le t}\Delta X_s$ if and only if $X$ is the sum of a continuous local martingale and an (absolutely convergent) sum of jump terms. This is really just tautological and saying that the identity holds only when it holds, but I doubt you are going to get any better general statement than that.
You also mention Hypothesis A: $\sum_{s\le t}\vert \Delta X_s\vert < \infty$ (a.s.). This property is not true for all semimartingales. In general, the existance of the quadratic variation ensures that
$$
\sum_{s\le t}(\Delta X_s)^2\le[X]_t < \infty.
$$
This does not imply Hypothesis A. You do have the inequality $\sum_{s\le t}\vert\Delta X_s\vert\ge\sqrt{\sum_{s\le t}(\Delta X_s)^2}$, but that is the wrong way round to tell us anything useful here. Examples of semimartingales which fail this hypothesis include the Cauchy process (I can link you to my blog post on properties of Levy processes for this, although any book on Levy processes should give more details), which are purely discontinuous semimartingales with finite variation.
Q2: This question has a very nice answer, $X^d$ is a finite variation process if and only if Hypothesis A holds! This can be proven without too much difficulty (assuming some facts about continuous semimartingales).
Proof: Suppose that $X^d$ is a finite variation process. As $\Delta X_t=\Delta X^d_t$, the sum $\sum_{s\le t}\vert\Delta X_s\vert$ is bounded by the variation of $X^d$ on the interval $[0,t]$, so is finite (a.s.)
In the other direction, suppose that Hypothesis A holds, so we can define the finite variation process $Y_t=\sum_{s\le t}\Delta X_s$. Then $X-Y$ is a continuous semimartingale. By a standard result on decomposition of continuous semimartingales, we can decompose $X-Y=M+V$ for a continuous local martingale $M$ and continuous finite variation process $V$. Then, $Y+V$ is a finite variation process, so is a purely discontinuous semimartingale. Also, $X=M+(Y+V)$ implies that $X^d=Y+V$ is a finite variation process.
