The canonical baseband version of BPSK is that we wish to transmit a sequence $\{a_k\}$ of bits ($0$ or $1$) at a rate of $1$ bit every $T$ seconds using a pulse $g(t)$ of duration $T$ and so the transmitted baseband signal is
$$s_{\text{baseband}}(t) = \sum_{k=-\infty}^\infty (-1)^{a_k}g(t-kT).$$
This is a sequence of nonoverlapping pulses of the form $\pm g(\cdot)$ spaced $T$ seconds apart. If, for example, $g(t)$ has value $1$ for $t \in [0,T)$, then $s(t)$ is a sequence of rectangular pulses of duration $T$
with the $k$-th pulse lasting from $t=kT$ to
$t=(k+1)T$. Let's use this specific $g(t)$ for the rest of this answer. Note that $s_{\text{baseband}}(t)$ has value $0$ or $1$ for all $t$, and that it changes from one value to the other at times tenant are multiples of the bit duration $T$. Some elementary books on digital communication extend these notions to the passband signal too, saying that the passband BPSK signal during $[0,T)$ is of the form (say) $\pm g(t)\cos(2\pi f_ct)$ (where $f_c$ is the carrier frequency) and so during $[T,2T)$, the passband BPSK signal is just this passband RF burst delayed in time by $T$ seconds. That is, such texts take the passband BPSK signal to be
$$s_{\text{passband}}(t) = \sum_{k=-\infty}^\infty (-1)^{a_k}g(t-kT)\cos(2\pi f_c(t-kT)).$$
Such a re-setting of the carrier phase to $0$ every $T$ seconds leads to discontinuities such as seen in the OP's first figure which correspond to $a_0=a_1 =1$.
In practice, and even in more advanced textbooks on digital communications, the BPSK passband signal is of the form
$$s_{\text{passband}}(t) = s_{\text{baseband}}(t)\cos(2\pi f_ct) = \left(\sum_{k=-\infty}^\infty (-1)^{a_k}g(t-kT)\right)\cos(2\pi f_ct)$$
which makes the modulation aspect very clear: there is an oscillator that is producing the carrier signal $\cos(2\pi f_ct)$, and this is being modulated by the baseband BPSK signal. Sure looks like amplitude modulation to the untrained eye, but it is indeed phase modulation, because the baseband signal has value $\pm 1$ and since $\cos(2\pi f_ct-\pi)=-\cos(2\pi f_ct)$, we can write
$$s_{\text{passband}}(t) = \cos\left(2\pi f_ct - \pi\sum_{k=-\infty}^\infty a_kg(t-kT)\right)$$
where that inner sum has value $1$ or $0$ for all $t$ and so $s_{\text{passband}}(t)$ is a sinusoidal signal whose phase at all times is either $0$ or $\pi$, and the phase can flip from $0$ to $\pi$ or vice versa only at $T$ second intervals. With $a_0=a_1=1$, the phase remains at $\pi$ during $[0,2T)$ leading to the second plot provided by the OP.
