It seems that you essentially have all the ingredients in place except for a couple of definitions.
Weights of $\mathbb{C}^*$: An action of $\mathbb C^*$ on $\mathbb{C}^n$ is defined by a group homomorphism: $\rho:\mathbb{C}^* \to \mathrm{Aut}(\mathbb{C}^n) = GL_n$. Since $\mathbb{C}^*$ is an abelian group, its irreducible representations are all one dimensional. So let's consider the representation $\rho: \mathbb{C}^* \to GL_1 = \mathbb{C}^*$, i.e., for any $z \in \mathbb{C}^*$, $\rho(z)$ is a nonzero complex number. We can use a Laurent series around $z=0$ to write $\rho(z)$:
$$ \rho(z) = \sum_{k=-\infty}^\infty a_k z^k\,. \tag{L}$$
Since $\rho$ is a group homomorphism, it has to satisfy:
$$\rho(z) \rho(w) = \rho(wz)\,.$$
If we use the expansion (L) on both sides of the above equation, we find:
$$ \sum_{k=-\infty}^\infty \sum_{l=-\infty}^\infty a_k a_l w^k z^l = \sum_{k=-\infty}^\infty a_k (wz)^k\,. $$
Since this has to be satisfied for all $w, z \in \mathbb{C}^*$, we must impose that $a_k$ is zero for all but one integer value of $k$ and if $a_k \ne 0$ then we must have $a_k=1$, i.e., irreducible representations of $\mathbb{C}^*$ can be parametrized by integers, and they are all of the form:
$$\rho_k(z) = z^k\,, \qquad k \in \mathbb Z\,.$$
This number $k$ is called the weight of the representation.
So, when it's said that the weight of the $\mathbb{C}^*$-action on $\mathbb{C}^{m+1}$ is $(\lambda_0, \cdots, \lambda_m)$, it's implied that the action is described as follows: for any $z \in \mathbb{C}^*$ and $(v_0, \cdots, v_m) \in \mathbb C^{m+1}$:
$$\rho(z): (v_0, \cdots, v_m) \mapsto (z^{\lambda_0} v_0, \cdots, z^{\lambda_m} v_m)\,.$$
Induced action: I am not going to give a formal definition of the induced action, rather I will give a "computational" definition. But if you haven't seen it already in texts then I suggest that you read it from a proper text (I am familiar with the one by Nakahara (Geometry, Topology and Physics) but surely there are many options).
Suppose we are in a local chart on some (locally) complex manifold $M$ with a group action. So we have local coordinates $\mathbf{x}=(x^1, \cdots, x^d)$ and the group action is given as a coordinate transformation:
$$g:(x^1, \cdots, x^d) \mapsto (g^1(\mathbf x), \cdots, g^d(\mathbf x)) =: g(\mathbf x). \tag{G}$$
At any point $\mathbf p = (p^1, \cdots, p^d) $ in this local chart, the tangent space $T_\mathbf{p} M$ is spanned by the partial derivatives at that point:
$$ T_\mathbf{p}M = \mathrm{Span}_\mathbb{C} \left( \left.\frac{\partial}{\partial x^1}\right|_{\mathbf{x} = \mathbf{p}}\,, \cdots, \left.\frac{\partial}{\partial x^d}\right|_{\mathbf{x} = \mathbf{p}} \right)\,. $$
Rcall from chain rule that the transformation (G) induces the following transformation for partial derivative:
$$ \left.\frac{\partial}{\partial x^i}\right|_{\mathbf{x} = \mathbf{p}} \mapsto \sum_{j=1}^d \left.\frac{\partial g^j(\mathbf x)}{\partial x^i}\right|_{\mathbf{x} = \mathbf{p}} \left.\frac{\partial}{\partial x^j}\right|_{\mathbf{x} = \mathbf{g(p)}}\,. \tag{T}$$
Noting that $\left.\frac{\partial}{\partial x^i}\right|_{\mathbf{x} = \mathbf{g(p)}}$ are basis vectors of $T_{g(\mathbf p)}M$, we thus get a map (which I denote here by $g_*$):
$$ g_*(\mathbf p): T_\mathbf{p}M \to T_{g(\mathbf p)}M\,.$$
Let us define the following matrix:
$$J(\mathbf p)_i^j := \left.\frac{\partial g^j(\mathbf x)}{\partial x^i}\right|_{\mathbf{x} = \mathbf{p}}. \tag{J}$$
Now, if we want to write down the matrix form of the linear transformation (T) then we should write the new basis vectors $\left\{\left.\frac{\partial}{\partial x^i}\right|_{\mathbf{x} = \mathbf{g(p)}}\right\}$ in terms of the old basis vectors $\left\{\left.\frac{\partial}{\partial x^i}\right|_{\mathbf{x} = \mathbf{p}}\right\}$, inverting the chain rule we get:
$$ \left. \frac{\partial}{\partial x^i} \right|_{\mathbf x = \mathbf g(p)} = \sum_{j=1}^d \left(J(\mathbf p)^{-1}\right)_i^j \left. \frac{\partial}{\partial x^j} \right|_{\mathbf x = \mathbf p}\,. $$
Therefore, $J(\mathbf p)^{-1}$ is the matrix form of the induced action on the tangent spaces, given an action on a manifold. If $\mathbf p$ happens to be a fixed point of the group action then we have a linear representation of the group $g_*(\mathbf p):T_\mathbf{p}M \to T_\mathbf{p} M$.
Question 1: The weights of the $\mathbb C^*$ action (let's denote the action by $\rho^{\mathbb C^{m+1}}$) on $\mathbb C^{m+1}$ is $(\lambda_0, \cdots, \lambda_m)$, which means, for any $z \in \mathbb{C}^*$ and $(x_0, \cdots, x_m) \in \mathbb C^{m+1}$ we have:
$$ \rho^{\mathbb C^{m+1}}(z):(x_0, \cdots, x_m) \mapsto (z^{\lambda_0} x_0, \cdots, z^{\lambda_m} x_m)\,.$$
Then your own computation shows that, on the local chart $U_0$ of $\mathbb P^m$, the action of $\mathbb C^*$ (let's denote it by $\rho^{\mathbb{P}^m}$) is given by:
$$\rho^{\mathbb P^m}(z):(y_1, \cdots, y_m) \mapsto (z^{\lambda_1-\lambda_0} y_1, \cdots, z^{\lambda_m-\lambda_0} y_m)\,.$$
This is the analogue of the coordinate transformation (G) and therefore we have $\left(\rho^{\mathbb P^m}(z)\right)^i (\mathbf y) = z^{\lambda_i-\lambda_0} y_i$ and the inverse of the matrix $J$ (defined in (J)) is given by:
$$J^{-1}(\mathbf y) = \mathrm{diag}(z^{\lambda_0 -\lambda_1}, \cdots, z^{\lambda_0 -\lambda_m})\,.$$
The induced action on tangent spaces is just multiplication by the above matrix, so we see that the weights are given by $(\lambda_0-\lambda_1, \cdots, \lambda_0 - \lambda_m)$. (Note that it's a representation only at the fixed point $\mathbf y = 0$, otherwise it's a linear map between two different vector spaces.)
Question 2: If you have the action of $(\mathbb C^*)^m$ on $\mathbb C^n$ then for each factor of $\mathbb C$ you will need an $m$-tuple of weights, let's say if we denote the $m$-tuple of weights corresponding to the $a$'th factor of $\mathbb C$ by $(w^a_1, \cdots, w^a_m)$, then we will write the action of $(t_1, \cdots, t_m) \in (\mathbb C^*)^m$ on $(x_1, \cdots, x_n) \in \mathbb C^n$ as:
$$ \rho(t_1, \cdots, t_m): (x_1, \cdots, x_n) \mapsto (t_1^{w^1_1} \cdots t_m^{w^1_m} x_1, \cdots, t_1^{w^n_1} \cdots t_m^{w^n_m} x_n)\,. $$
