Denoting the current day as index $i$ (so that yesterday is $i-1$ and the day before yesterday is $i-2$), the previous state comprises two elements, $State_{n-1}=\{S_{i-1},S_{i-2}\}$, where $S_k$ is the state of the weather for day $k$. Thus the current state is $State_{n}=\{S_{i},S_{i-1}\}$
Lettting $R$ be the occurrence of rain on a particular day, so that $\overline{R}$ is when no rain occurs for that day, there are $4$ possible values of the previous state $State_{n-1}$
$$\{\overline{R},\overline{R}\}, \{\overline{R},R\},\{R,\overline{R}\},\{R,R\}$$
For each of these $4$ states, there are only two possible next states:-
$$\begin{align}
State_n=\{\overline{R},\overline{R}\}\text { or }State_n=\{R,\overline{R}\}\text{ given }State_{n-1}=\{\overline{R},\overline{R}\}
\\
State_n=\{\overline{R},\overline{R}\}\text { or }State_n=\{R,\overline{R}\}\text{ given }State_{n-1}=\{\overline{R},R\}
\\
State_n=\{\overline{R},R\}\text { or }State_n=\{R,R\}\text{ given }State_{n-1}=\{R,\overline{R}\}
\\
State_n=\{\overline{R},R\}\text { or }State_n=\{R,R\}\text{ given }State_{n-1}=\{R,R\}
\end{align}$$
With the information given in the question, and based on the constraints of what the next state is based on the current state, you can figure out the structure of the transition matrix.
For example, the information Specifically, suppose that if it has rained for the past two days, then it will rain tomorrow with probability $0.7$, corresponds to
$$P(State_{n+1}=\{\overline{R},\overline{R}\}|State_{n}=\{\overline{R},\overline{R}\})=0.7\\\Rightarrow P(State_{n+1}=\{R,\overline{R}\}|State_{n}=\{\overline{R},\overline{R}\})=1-0.7=0.3$$
and if it rained today but not yesterday, then it will rain tomorrow with probability 0.5 corresponds to
$$P(State_{n+1}=\{\overline{R},\overline{R}\}|State_{n}=\{\overline{R},R\})=0.5\\\Rightarrow P(State_{n+1}=\{R,\overline{R}\}|State_{n}=\{\overline{R},R\})=1-0.5=0.5$$
Alice carrying an umbrella on day $\ d\ $ is the same event as that the Markov chain is in either state $\ 2\ $ or state $\ 3\ $ on day $\ d\ $. So the probability that Alice is carrying an umbrella today, given that she carried one yesterday is
$$
P\big(X_0=3\big| X_{-1}=3\ \text{ or }\ X_{-1}=4 \big)+ P\big(X_0=4\big| X_{-1}=3\ \text{ or }\ X_{-1}=4 \big)\\
=\frac{P\big(X_0=3,X_{-1}=3\big)+ P\big(X_0=3,X_{-1}=4\big)}{P\big(X_{-1}=3\big)+ P\big(X_{-1}=4\big)}\\
+ \frac{P\big(X_0=4,X_{-1}=3\big)+ P\big(X_0=4,X_{-1}=4\big)}{P\big(X_{-1}=3\big)+ P\big(X_{-1}=4\big)}\\
=\frac{\displaystyle\sum_{i=3}^4\sum_{j=3}^4 \pi(-1)_i P_{ij}}{\pi(-1)_3+\pi(-1)_4}\ ,
$$
Similarly, the probability that Alice is carrying an umbrella today, given that she carried one on the last two days is
$$
\frac{\displaystyle\sum_{i=3}^4\sum_{j=3}^4 \sum_{k=3}^4\pi(-2)_i P_{ij}P_{jk}}{\displaystyle\sum_{i=3}^4\sum_{j=3}^4\pi(-2)_iP_{ij}}\ .
$$
To evaluate these quantities, you need to know the state distribution on the preceding two days, which you can presumably assume to be the stationary distribution. Your intuition that these two quantities should be equal, however, has led you astray. They aren't.
Best Answer
\begin{matrix}&RR&SR&RS&SS\\RR&0.7&0&0.3&0\\SR&0.5&0&0.5&0\\RS&0&0.4&0&0.6\\SS&0&0.2&0&0.8\\ \end{matrix} You start by the row and end up by the column:
If the first condition is RR (rained yesterday and rained tomorrow), the probability of RR (rained today and will rain tomorow) is 0.7 or RS (rained today dry tomorow) 0.3.
Note that every row sums up to 1.0