Here is what they mean: For any $m\in\mathbb{N}$, there is a canonical
embedding
\begin{align*}
\iota_{\operatorname*{Sym},m}:\operatorname*{Sym}\nolimits^{m}V & \rightarrow
V^{\otimes m},\\
v_{1}v_{2}\cdots v_{m} & \mapsto\sum_{\sigma\in S_{m}}v_{\sigma\left(
1\right) }\otimes v_{\sigma\left( 2\right) }\otimes\cdots\otimes
v_{\sigma\left( m\right) }
\end{align*}
(where the "$v_{1}v_{2}\cdots v_{m}$" on the left hand side means the
projection of the tensor $v_{1}\otimes v_{2}\otimes\cdots\otimes v_{m}\in
V^{\otimes m}$ onto $\operatorname*{Sym}\nolimits^{m}V$). Thus, you have
canonical embeddings $\iota_{\operatorname*{Sym},\lambda_{i}}
:\operatorname*{Sym}\nolimits^{\lambda_{i}}V\rightarrow V^{\otimes\lambda_{i}
}$ for all $i\in\left\{ 1,2,\ldots,k\right\} $. The tensor product of these
$k$ embeddings is an embedding
\begin{align*}
\iota_{\operatorname*{Sym},\lambda_{1}}\otimes\iota_{\operatorname*{Sym}
,\lambda_{2}}\otimes\cdots\otimes\iota_{\operatorname*{Sym},\lambda_{k}} &
:\operatorname*{Sym}\nolimits^{\lambda_{1}}V\otimes\operatorname*{Sym}
\nolimits^{\lambda_{2}}V\otimes\cdots\otimes\operatorname*{Sym}
\nolimits^{\lambda_{k}}V\\
& \rightarrow V^{\otimes\lambda_{1}}\otimes V^{\otimes\lambda_{2}}
\otimes\cdots\otimes V^{\otimes\lambda_{k}}.
\end{align*}
If we identify $V^{\otimes\lambda_{1}}\otimes V^{\otimes\lambda_{2}}
\otimes\cdots\otimes V^{\otimes\lambda_{k}}$ with $V^{\otimes n}$ in the
standard way (i.e., identifying each tensor
\begin{align*}
& \left( a_{1}\otimes a_{2}\otimes\cdots\otimes a_{\lambda_{1}}\right)
\otimes\left( b_{1}\otimes b_{2}\otimes\cdots\otimes b_{\lambda_{2}}\right)
\otimes\cdots\otimes\left( g_{1}\otimes g_{2}\otimes\cdots\otimes g_{\lambda_{k}
}\right) \\
& \in V^{\otimes\lambda_{1}}\otimes V^{\otimes\lambda_{2}}\otimes\cdots\otimes
V^{\otimes\lambda_{k}}
\end{align*}
(where the letters "$a$" and "$b$" here have nothing to do with $a_{\lambda}$
and $b_{\lambda}$) with
\begin{align*}
& a_{1}\otimes a_{2}\otimes\cdots\otimes a_{\lambda_{1}}\otimes b_{1}\otimes
b_{2}\otimes\cdots\otimes b_{\lambda_{2}}\otimes\cdots\otimes g_{1}\otimes
g_{2}\otimes\cdots\otimes g_{\lambda_{k}}\\
& \in V^{\otimes n}
\end{align*}
), then this becomes an embedding
\begin{equation}
\operatorname*{Sym}\nolimits^{\lambda_{1}}V\otimes\operatorname*{Sym}
\nolimits^{\lambda_{2}}V\otimes\cdots\otimes\operatorname*{Sym}
\nolimits^{\lambda_{k}}V\rightarrow V^{\otimes n}.
\end{equation}
Fulton/Harris regard this embedding as an inclusion (another abuse of
notation), i.e., they use it to pretend that $\operatorname*{Sym}
\nolimits^{\lambda_{1}}V\otimes\operatorname*{Sym}\nolimits^{\lambda_{2}
}V\otimes\cdots\otimes\operatorname*{Sym}\nolimits^{\lambda_{k}}V$ is a
$\mathbb{C}$-vector subspace of $V^{\otimes n}$. Now they claim that the image
$\operatorname{Im}\left( a_{\lambda}\right) $ is precisely this subspace.
This yields, in particular, your claim that anything in this image is
unchanged when you hit it with a permutation $\sigma\in S_{\left\{
\lambda_{\leq i}+1,\ldots,\lambda_{\leq i+1}\right\} }$; but it is a stronger statement.
Why is it true? Well, it isn't true in general. You need to assume that the
tableau you are using to define $a_{\lambda}$ is the one whose first row has
entries $1,2,\ldots,\lambda_{1}$ (in this order), whose second row has entries
$\lambda_{1}+1,\lambda_{1}+2,\ldots,\lambda_{1}+\lambda_{2}$ (in this order),
and so on (i.e., if you read it row by row from top to bottom, then you get
the sequence $\left( 1,2,\ldots,n\right) $). Then, it is easy to see that
the permutations $g\in P_{\lambda}$ are exactly the permutations in $S_{n}$
that can be written in the form $\sigma_{1}\sigma_{2}\cdots\sigma_{k}$, where
each $\sigma_{i}$ belongs to $S_{\left\{ \lambda_{\leq i-1}+1,\lambda_{\leq
i-1}+2,\ldots,\lambda_{\leq i}\right\} }$. Moreover, they can be written in
this form uniquely; thus, $P_{\lambda}$ is the following internal direct
product of subgroups of $S_{n}$:
\begin{equation}
P_{\lambda}=\prod_{i=1}^{k}S_{\left\{ \lambda_{\leq i-1}+1,\lambda_{\leq
i-1}+2,\ldots,\lambda_{\leq i}\right\} }.
\end{equation}
Hence, every $v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\in V^{\otimes n}$
satisfies
\begin{align*}
& \sum_{g\in P_{\lambda}}g\left( v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{n}\right) \\
& =\sum_{\substack{\left( \sigma_{1},\sigma_{2},\ldots,\sigma_{k}\right)
;\\\text{each }\sigma_{i}\text{ belongs to }S_{\left\{ \lambda_{\leq
i-1}+1,\lambda_{\leq i-1}+2,\ldots,\lambda_{\leq i}\right\} }}
}\underbrace{\left( \sigma_{1}\sigma_{2}\cdots\sigma_{k}\right) \left(
v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right) }_{\substack{=\bigotimes
_{i=1}^{k}\sigma_{i}\left( v_{\lambda_{\leq i-1}+1}\otimes v_{\lambda_{\leq
i-1}+2}\otimes\cdots\otimes v_{\lambda_{\leq i}}\right) \\\text{(where the
notation }\bigotimes_{i=1}^{k}w_{i}\text{ is shorthand for }w_{1}\otimes
w_{2}\otimes\cdots\otimes w_{k}\text{,}\\\text{whenever }w_{1}\in
V^{\otimes\lambda_{1}},w_{2}\in V^{\otimes\lambda_{2}},\ldots,w_{k}\in
V^{\otimes\lambda_{k}}\text{)}}}\\
& =\sum_{\substack{\left( \sigma_{1},\sigma_{2},\ldots,\sigma_{k}\right)
;\\\text{each }\sigma_{i}\text{ belongs to }S_{\left\{ \lambda_{\leq
i-1}+1,\lambda_{\leq i-1}+2,\ldots,\lambda_{\leq i}\right\} }}}\bigotimes
_{i=1}^{k}\sigma_{i}\left( v_{\lambda_{\leq i-1}+1}\otimes v_{\lambda_{\leq
i-1}+2}\otimes\cdots\otimes v_{\lambda_{\leq i}}\right) \\
& =\bigotimes_{i=1}^{k}\underbrace{\sum_{\sigma_{i}\in S_{\left\{
\lambda_{\leq i-1}+1,\lambda_{\leq i-1}+2,\ldots,\lambda_{\leq i}\right\} }
}\sigma_{i}\left( v_{\lambda_{\leq i-1}+1}\otimes v_{\lambda_{\leq i-1}
+2}\otimes\cdots\otimes v_{\lambda_{\leq i}}\right) }_{=\iota
_{\operatorname*{Sym},\lambda_{i}}\left( v_{\lambda_{\leq i-1}+1}\otimes
v_{\lambda_{\leq i-1}+2}\otimes\cdots\otimes v_{\lambda_{\leq i}}\right) }\\
& =\bigotimes_{i=1}^{k}\iota_{\operatorname*{Sym},\lambda_{i}}\left(
v_{\lambda_{\leq i-1}+1}\otimes v_{\lambda_{\leq i-1}+2}\otimes\cdots\otimes
v_{\lambda_{\leq i}}\right) \\
& =\left( \iota_{\operatorname*{Sym},\lambda_{1}}\otimes\iota
_{\operatorname*{Sym},\lambda_{2}}\otimes\cdots\otimes\iota
_{\operatorname*{Sym},\lambda_{k}}\right) \left( v_{1}\otimes v_{2}
\otimes\cdots\otimes v_{n}\right) .
\end{align*}
Since $\sum_{g\in P_{\lambda}}g=a_{\lambda}$, this rewrites as
\begin{equation}
a_{\lambda}\left( v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right)
=\left( \iota_{\operatorname*{Sym},\lambda_{1}}\otimes\iota
_{\operatorname*{Sym},\lambda_{2}}\otimes\cdots\otimes\iota
_{\operatorname*{Sym},\lambda_{k}}\right) \left( v_{1}\otimes v_{2}
\otimes\cdots\otimes v_{n}\right) .
\end{equation}
Thus, the image $\operatorname*{Im}\left( a_{\lambda}\right) $ is the image
of the map $\iota_{\operatorname*{Sym},\lambda_{1}}\otimes\iota
_{\operatorname*{Sym},\lambda_{2}}\otimes\cdots\otimes\iota
_{\operatorname*{Sym},\lambda_{k}}$. Since we have agreed to identify the
latter image with $\operatorname*{Sym}\nolimits^{\lambda_{1}}V\otimes
\operatorname*{Sym}\nolimits^{\lambda_{2}}V\otimes\cdots\otimes
\operatorname*{Sym}\nolimits^{\lambda_{k}}V$, we thus conclude that
\begin{equation}
\operatorname{Im}\left( a_{\lambda}\right) =\operatorname*{Sym}
\nolimits^{\lambda_{1}}V\otimes\operatorname*{Sym}\nolimits^{\lambda_{2}
}V\otimes\cdots\otimes\operatorname*{Sym}\nolimits^{\lambda_{k}}V.
\end{equation}
This is similar to part 2. Instead of the embeddings $\iota
_{\operatorname*{Sym},m}$, you now need the embeddings
\begin{align*}
\iota_{\Lambda,m}:\Lambda^{m}V & \rightarrow V^{\otimes m},\\
v_{1}\wedge v_{2}\wedge\cdots\wedge v_{m} & \mapsto\sum_{\sigma\in S_{m}
}\left( -1\right) ^{\sigma}v_{\sigma\left( 1\right) }\otimes
v_{\sigma\left( 2\right) }\otimes\cdots\otimes v_{\sigma\left( m\right) }
\end{align*}
(where $\left( -1\right) ^{\sigma}$ is the sign of $\sigma$). These give you
canonical embeddings $\iota_{\Lambda,\mu_{i}}:\Lambda^{\mu_{i}}V\rightarrow
V^{\otimes\mu_{i}}$ for all $i\in\left\{ 1,2,\ldots,l\right\} $. The tensor
product of these $l$ embeddings is an embedding
\begin{align*}
\iota_{\Lambda,\mu_{1}}\otimes\iota_{\Lambda,\mu_{2}}\otimes\cdots\otimes
\iota_{\Lambda,\mu_{l}} & :\Lambda^{\mu_{1}}V\otimes\Lambda^{\mu_{2}}
V\otimes\cdots\otimes\Lambda^{\mu_{l}}V\\
& \rightarrow V^{\otimes\mu_{1}}\otimes V^{\otimes\mu_{2}}\otimes\cdots\otimes
V^{\otimes\mu_{l}}.
\end{align*}
Identifying $V^{\otimes\mu_{1}}\otimes V^{\otimes\mu_{2}}\otimes\cdots\otimes
V^{\otimes\mu_{l}}$ with $V^{\otimes n}$ in the standard way, this becomes an
embedding
\begin{equation}
\Lambda^{\mu_{1}}V\otimes\Lambda^{\mu_{2}}V\otimes\cdots\otimes\Lambda
^{\mu_{l}}V\rightarrow V^{\otimes n}.
\end{equation}
Fulton/Harris claim that the image $\operatorname{Im}\left( b_{\lambda
}\right) $ is precisely the image of this embedding.
This is, again, not true in general. This time, you need to assume that the
tableau you are using to define $b_{\lambda}$ is the one whose first
column has entries $1,2,\ldots,\mu_{1}$ (in this order), whose second
column has entries $\mu_{1}+1,\mu_{1}+2,\ldots,\mu_{1}+\mu_{2}$ (in this
order), and so on (i.e., if you read it column by column from left to right,
then you get the sequence $\left( 1,2,\ldots,n\right) $). The proof of this
is similar to the proof of 2.
This means that the claim in 2. and the claim 3. cannot hold at the same time:
The tableau that is required for 2. and the tableau that is required for 3.
are different (unless $\lambda$ consists merely of a single row or of a single
column). My impression is that Fulton/Harris fix this problem by identifying
$V^{\otimes\mu_{1}}\otimes V^{\otimes\mu_{2}}\otimes\cdots\otimes
V^{\otimes\mu_{l}}$ with $V^{\otimes n}$ in the non-standard way: instead of
just "dropping the parentheses", they permute the tensor factors so that the
first tensor factors in each parentheses are read first, then the second
tensor factors, etc. In other words, they identify each tensor
\begin{align*}
& \left( a_{1}\otimes a_{2}\otimes\cdots\otimes a_{\mu_{1}}\right)
\otimes\left( b_{1}\otimes b_{2}\otimes\cdots\otimes b_{\mu_{2}}\right)
\otimes\cdots\otimes\left( g_{1}\otimes g_{2}\otimes\cdots\otimes g_{\mu_{l}
}\right) \\
& \in V^{\otimes\mu_{1}}\otimes V^{\otimes\mu_{2}}\otimes\cdots\otimes
V^{\otimes\mu_{l}}
\end{align*}
(where the letters "$a$" and "$b$" here have nothing to do with $a_{\lambda}$
and $b_{\lambda}$) with
\begin{align*}
& \underbrace{a_{1}\otimes b_{1}\otimes\cdots}_{\text{all tensor factors with
subscript }1}\otimes\underbrace{a_{2}\otimes b_{2}\otimes\cdots}_{\text{all
tensor factors with subscript }2}\otimes\cdots\otimes\underbrace{a_{\mu_{1}
}\otimes b_{\mu_{1}}\otimes\cdots}_{\text{all tensor factors with subscript
}\mu_{1}}\\
& \in V^{\otimes n},
\end{align*}
instead of with the tensor
\begin{align*}
& a_{1}\otimes a_{2}\otimes\cdots\otimes a_{\mu_{1}}\otimes b_{1}\otimes
b_{2}\otimes\cdots\otimes b_{\mu_{2}}\otimes\cdots\otimes g_{1}\otimes
g_{2}\otimes\cdots\otimes g_{\mu_{l}}\\
& \in V^{\otimes n}
\end{align*}
as we did above. If you make this change, then the tableau that is required
for 3. becomes the same as that required for 2.
