The graph in your first question has no perfect matching. Since there are only $3$ vertices in $P_1$, a matching using all of those vertices can use at most $3$ of the $7$ vertices of $P_2$ and therefore cannot use all of them.
The two definitions are indeed of two different things. A perfect matching exhausts all of the vertices, so a bipartite graph that has a perfect matching must have the same number of vertices in each part. Deo is defining a directional notion: a complete matching from one part into the other. If $P_1$ and $P_2$ are the parts of a bipartite graph, $P_1$ had $m$ vertices, and $P_2$ has $n$ vertices, then a complete matching of $P_1$ into $P_2$ has $m$ edges, and a complete matching of $P_2$ into $P_1$ has $n$ edges. Thus, the former is possible only if $m\le n$, the latter is possible only if $n\le m$, and a perfect matching (which is automatically a complete matching of $P_1$ into $P_2$ and a complete matching of $P_2$ into $P_1$) is possible only if $m=n$.
Let $V = \{v_i\}_{1\leq i \leq n}$ and $W = \{w_j\}_{1\leq j \leq m}$ be sets of $n$ and $m$ vertices corresponding to each part. I will write $[\![s]\!]$ to refer to $\{1,\dots ,s\}$. Essentially, we can characterize each bipartite graph $G$ with a function
$$
\Gamma_G : [\![n]\!] \longrightarrow \mathcal{P}([\![m]\!]) \setminus \{\emptyset\}
$$
such that $\Gamma_G(i) = \{j : w_j \in N(v_i)\}$. Note that from this we can also recover $N(w_j)$ for $w_j \in W$. Therefore, our problem reduces to counting the amount functions from $[\![n]\!]$ to $\mathcal{P'} := \mathcal{P}([\![m]\!]) \setminus \{\emptyset\}$ such that every vertex in $W$ has a neighbour, that is, $\cup_{i \in [\![n]\!]} \Gamma(i) = [\![m]\!]$. I will denote this set as
$$
\mathcal{F} = \{\Gamma \in (\mathcal{P}')^{[\![n]\!]} : \cup_{i \in [\![n]\!]} \Gamma(i) = [\![m]\!]\}
$$
Since there are $(2^m -1)^n$ functions of the form $f : [\![n]\!] \to \mathcal{P}'$ in total, it would suffice to count $\mathcal{F}^c$ and substract it from $(2^m -1)^n$. Now, because
$$
\mathcal{F}^c = \bigcup_{j \in [\![m]\!]}\{\Gamma \in (\mathcal{P}')^{[\![n]\!]} : j \not \in \Gamma(i) \ (\forall i \in [\![n]\!])\}
$$
we can count this set using the inclusion-exclusion principle:
$$
|\mathcal{F}^c| = \sum_{k = 1}^m(-1)^{k+1}\sum_{\emptyset \subsetneq S \subseteq [\![m]\!], |S| = k}|\bigcap_{s \in S}\{\Gamma \in (\mathcal{P}')^{[\![n]\!]} : s \not \in \Gamma(i) \ (\forall i \in [\![n]\!])\}|
$$
More concisely,
$$
|\mathcal{F}^c| = \sum_{k = 1}^m(-1)^{k+1}\sum_{\emptyset \subsetneq S \subseteq [\![m]\!], |S| = k}|\{\Gamma \in (\mathcal{P}')^{[\![n]\!]} : S \cap \Gamma(i) = \emptyset \ (\forall i \in [\![n]\!])\}|
$$
Let's count each set of the inner sum. If $S$ is a proper subset of $[\![m]\!]$, to give a function such that each image is disjoint with $S$ is the same as giving a function that maps each $i \in [\![n]\!]$ to a subset of $[\![m]\!] \setminus S$ (except the empty set which we have already excluded). Therefore, for each $i \in [\![n]\!]$ we have $2^{m-|S|}-1$ choices, giving a total of $(2^{m-|S|}-1)^n$. Thus,
$$
|\mathcal{F}^c| = \sum_{k = 1}^m(-1)^{k+1}\sum_{\emptyset \subsetneq S \subseteq [\![m]\!], |S| = k}(2^{m-|S|}-1)^n = \sum_{k = 1}^m(-1)^{k+1}{m\choose k}(2^{m-k}-1)^n
$$
And finally,
$$
|\mathcal{F}| = |(\mathcal{P})^{[\![n]\!]} \setminus \mathcal{F}^c| = |(\mathcal{P})^{[\![n]\!]}| - |\mathcal{F}^c| = (2^m -1)^n - |\mathcal{F}^c| = \\=(2^m -1)^n + \sum_{k = 1}^m{m\choose k}(-1)^k(2^{m-k}-1)^n = \\
= \sum_{k = 0}^m{m\choose k}(-1)^k(2^{m-k}-1)^n = \sum_{k = 1}^m{m\choose k}(-1)^{m-k}(2^{k}-1)^n.
$$
Best Answer
In the product $Sx$, $x\in\mathbf{F}_2^{\lvert E\rvert}$, the vector $x$ corresponds to a set of edges of $G$. As you've observed, $Sx=0$ means that every vertex of $G$ is incident on an even number of edges of the edge set represented by $x$. The set of edges of a cycle of $G$ has this property. More generally, the set of edges of tour of $G$ has this property. (In a tour, vertices may be visited multiple times, but edges are used only once.) More generally still, the set of edges of a disjoint union of tours has this property. Such disjoint unions are equivalent to edge sets of even-degree subgraphs, or, in other words, subgraphs in which every connected component has an Euler tour.
With regard to the row space of $S$: in the product $x^TS$, $x\in\mathbf{F}_2^{\lvert V\rvert}$, the vector $x$ represents a set of vertices of $G$. The elements of the vector $x^TS$ correspond to edges. As you have found, an element of $x^TS$ equals $0$ when either both vertices on the corresponding edge are in the vertex set corresponding to $x$ or neither vertex is in that set; an element of $x^TS$ equals $1$ when exactly one of the vertices on the corresponding edge is in the vertex set corresponding to $x$. Hence the vector $x^TS$ represents the set of edges that join a vertex in the set corresponding to $x$ to a vertex not in that set. There is an induced subgraph corresponding to $x$, and $x^TS$ can be thought of as the set of external edges—edges connecting to vertices outside the induced subgraph.
The correspondence between row-space vectors and sets of external edges of induced subgraphs is not one-to-one for the following reasons: (1) if $X$ is the sets of vertices corresponding to $x$, the the subgraph induced by $X$ has the same external edge set as the subgraph induced by $V\setminus X$; (2) if $G$ is not connected then any connected component or union of connected components has no external edges, just as the empty subgraph does. In other words, if $y_1$, $y_2$, ..., $y_m$ are vectors corresponding to vertex sets of connected components, then $y_j^TS=0$ for all $j\in\{1,2,\ldots,m\}$ and $$ (x^T+y_1^T+\ldots+y_m^T)S=x^TS $$ for any $x\in\mathbf{F}_2^{\lvert V\rvert}$.
So there is a one-to-one correspondence between vectors in the row space of $S$ and equivalence classes of partitions of the vertex set of $G$ into two parts, where the partition $\{X,Y\}$ is the same partition as $\{Y,X\}$. The notion of equivalence being used here is the following: let $V$ be partitioned as $\{X,Y,C\}$ where $C$ is the set of vertices of a connected component of $G$. Then $\{X\cup C,Y\}\sim\{X,Y\cup C\}$.